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Statistics and probability

Unit 8: lesson 3.

Permutations & combinations

COMBINATION PROBLEMS WITH SOLUTIONS

Problem 1 :

Find the number of ways in which 4 letters can be selected from the word ACCOUNTANT.

Problem 2 :

A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

Number of white balls  =  2

Number of black balls  =  3

Number of red balls  =  4

Number of non black balls  =  2 + 4  =  6

Number of ways 

  =  ( 3 C 1  ⋅  6 C 2 ) + ( 3 C 2  ⋅  6 C 1 )  +  ( 3 C 3  ⋅  6 C 0 )

  =  ( 3 ⋅ 15) +  ( 3 ⋅ 6) +  (1  ⋅ 1)

=  45 + 18 + 1

Problem 3 :

Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION?

There are 11 letters not all different.

They are AA, II, NN, E, X, M, T, O. The following combinations are possible: Case 1 :

Number of ways selecting 2 alike, 2 alike

  =   3 C 2    =  3 ways

Number of ways selecting 2 alike,2 different

  =   3 C 1  ⋅     7 C 2  ==>  3 x 21  ==>  63 ways.

Number of ways selecting all 4 different =    8 C 4  

Total number of combinations = 3 + 63 + 70 = 136 ways.

Total number of permutations (1) to (3)

  =  3  ⋅  (4!/2!2!) + 63  ⋅  (4!/2!) + 70  ⋅  4!

=  18 + 756 + 1680

Problem 4 :

How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?

From the given question, we come to know that any three points are not collinear.

By selecting any three points out of 15 points, we draw a triangle.

Number of ways to draw a triangle  =  15 C 3

  =  (15  ⋅  14  ⋅  13) / (3  ⋅ 2 ⋅ 1)  

  =  455

Problem 5 :

A committee of 7 members is to be chosen from 6 artists, 4 singers and 5 writers. In how many ways can this be done if in the committee there must be at least one member from each group and at least 3 artists ?

For the given condition, possible ways to select members for a committee of 7 members.

(3A, 3S, 1W) ----> 6C3 ⋅ 4C3 ⋅ 5C1  =  20 ⋅ 4 ⋅ 5 = 400

(3A, 1S, 3W) ----> 6C3 ⋅ 4C1 ⋅ 3C1  =  20 ⋅ 4 ⋅ 10 = 800

(3A, 2S, 2W) ----> 6C3 ⋅ 4C2 ⋅ 5C2 = 20 ⋅ 6 ⋅ 10 = 1200

(4A, 2S, 1W) ----> 6C4 ⋅ 4C2 ⋅ 5C1 = 15 ⋅ 6 ⋅ 5 = 450

(4A, 1S, 2W) ----> 6C4 ⋅ 4C1 ⋅ 5C2 = 15 ⋅ 4 ⋅ 10 = 600

(5A, 1S, 1W) ----> 6C5 ⋅ 4C1 ⋅ 5C1 = 6 ⋅ 4 ⋅ 5 = 120

Thus, the total no. of ways is

= 400 + 800 + 1200 + 450 + 600 + 120

Problem 6 :

The supreme court has given a 6 to 3 decisions upholding a lower court. Find the number of ways it can give a majority decision reversing the lower court.

Upholding a lower court means, supporting it for its decision.

Reversing a lower court means, opposing it for its decision.

In total of 9 cases (6 + 3 = 9), it may give 5 or 6 or 7 or 8 or 9 decisions reversing the lower court. And it can not be 4 or less than 4. Because majority of 9 is 5 or more.

The possible combinations in which it can give a majority decision reversing the lower court are

5 out of 9 ----> 9C5 = 126

6 out of 9 ----> 9C6 = 84

7 out of 9 ----> 9C7 = 36

8 out of 9 ----> 9C8 = 9

9 out of 9 ----> 9C9 = 1

Thus, the total number of ways is

= 126 + 84 + 36 + 9 + 1

Problem 7 :

Five bulbs of which three are defective are to be tried in two bulb points in a dark room. Find the number of trials in which the room can be lighted.

Given :  3 bulbs are defective out of 5.There are two bulb points in the dark room. One bulb (or two bulbs) in good condition is enough to light the room. Since there are two bulb points, we have to select 2 out of 5 bulbs. No. of ways of selecting 2 bulbs out of 5 is

(It includes selecting two good bulbs, two defective bulbs, one good bulb and one defective bulb. So, in these 10 ways, room may be lighted or may not be lighted)

Number of ways of selecting 2 defective bulbs out of 3 is

(It includes selecting only two defective bulbs. So, in these 3 ways, room can not be lighted)

The number of ways in which the room can be lighted is

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How to Calculate Combinations

Last Updated: June 10, 2021 References

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. There are 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 90,338 times. Learn more...

Permutations and combinations have uses in math classes and in daily life. Thankfully, they are easy to calculate once you know how. Unlike permutations , where group order matters, in combinations, the order doesn't matter. [1] X Research source Combinations tell you how many ways there are to combine a given number of items in a group. To calculate combinations, you just need to know the number of items you're choosing from, the number of items to choose, and whether or not repetition is allowed (in the most common form of this problem, repetition is not allowed).

Calculating Combinations Without Repetition

Image titled Calculate Combinations Step 1

{}_{{n}}C_{{r}}

Calculating Combinations with Repetition

{}_{{n+r-1}}C_{{r}}

{}_{{n+r-1}}C_{{r}}={\frac {(n+r-1)!}{(n-1)!r!}}

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Combinations and Permutations

What's the difference.

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.

"The combination to the safe is 472" . Now we do care about the order. "724" won't work, nor will "247". It has to be exactly 4-7-2 .

So, in Mathematics we use more precise language:

In other words:

A Permutation is an ordered Combination.

Permutations

There are basically two types of permutation:

1. Permutations with Repetition

These are the easiest to calculate.

When a thing has n different types ... we have n choices each time!

For example: choosing 3 of those things, the permutations are:

n × n × n (n multiplied 3 times)

More generally: choosing r of something that has n different types, the permutations are:

n × n × ... (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)

Which is easier to write down using an exponent of r :

n × n × ... (r times) = n r

Example: in the lock above, there are 10 numbers to choose from (0,1,2,3,4,5,6,7,8,9) and we choose 3 of them:

10 × 10 × ... (3 times) = 10 3 = 1,000 permutations

So, the formula is simply:

2. Permutations without Repetition

In this case, we have to reduce the number of available choices each time.

pool balls

Example: what order could 16 pool balls be in?

After choosing, say, number "14" we can't choose it again.

So, our first choice has 16 possibilites, and our next choice has 15 possibilities, then 14, 13, 12, 11, ... etc. And the total permutations are:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe we don't want to choose them all, just 3 of them, and that is then:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be arranged out of 16 balls.

Without repetition our choices get reduced each time.

But how do we write that mathematically? Answer: we use the " factorial function "

So, when we want to select all of the billiard balls the permutations are:

16! = 20,922,789,888,000

But when we want to select just 3 we don't want to multiply after 14. How do we do that? There is a neat trick: we divide by 13!

16 × 15 × 14 × 13 × 12 × ... 13 × 12 × ...   =  16 × 15 × 14

That was neat: the 13 × 12 × ... etc gets "cancelled out", leaving only 16 × 15 × 14 .

The formula is written:

Example Our "order of 3 out of 16 pool balls example" is:

(which is just the same as: 16 × 15 × 14 = 3,360 )

Example: How many ways can first and second place be awarded to 10 people?

(which is just the same as: 10 × 9 = 90 )

Instead of writing the whole formula, people use different notations such as these:

Combinations

There are also two types of combinations (remember the order does not matter now):

1. Combinations with Repetition

Actually, these are the hardest to explain, so we will come back to this later.

2. Combinations without Repetition

This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win!

The easiest way to explain it is to:

Going back to our pool ball example, let's say we just want to know which 3 pool balls are chosen, not the order.

We already know that 3 out of 16 gave us 3,360 permutations.

But many of those are the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 are chosen. These are the possibilites:

So, the permutations have 6 times as many possibilites.

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

3! = 3 × 2 × 1 = 6

(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)

So we adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in their order any more):

That formula is so important it is often just written in big parentheses like this:

It is often called "n choose r" (such as "16 choose 3")

And is also known as the Binomial Coefficient .

All these notations mean "n choose r":

Just remember the formula:

n! r!(n − r)!

Example: Pool Balls (without order)

So, our pool ball example (now without order) is:

16! 3!(16−3)!

= 16! 3! × 13!

= 20,922,789,888,000 6 × 6,227,020,800

Notice the formula 16! 3! × 13! gives the same answer as 16! 13! × 3!

So choosing 3 balls out of 16, or choosing 13 balls out of 16, have the same number of combinations:

16! 3!(16−3)! = 16! 13!(16−13)! = 16! 3! × 13! = 560

In fact the formula is nice and symmetrical :

Also, knowing that 16!/13! reduces to 16×15×14, we can save lots of calculation by doing it this way:

16×15×14 3×2×1

Pascal's Triangle

We can also use Pascal's Triangle to find the values. Go down to row "n" (the top row is 0), and then along "r" places and the value there is our answer. Here is an extract showing row 16:

OK, now we can tackle this one ...

ice cream

Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla .

We can have three scoops. How many variations will there be?

Let's use letters for the flavors: {b, c, l, s, v}. Example selections include

(And just to be clear: There are n=5 things to choose from, we choose r=3 of them, order does not matter, and we can repeat!)

Now, I can't describe directly to you how to calculate this, but I can show you a special technique that lets you work it out.

Think about the ice cream being in boxes, we could say "move past the first box, then take 3 scoops, then move along 3 more boxes to the end" and we will have 3 scoops of chocolate!

So it is like we are ordering a robot to get our ice cream, but it doesn't change anything, we still get what we want.

In fact the three examples above can be written like this:

So instead of worrying about different flavors, we have a simpler question: "how many different ways can we arrange arrows and circles?"

Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container).

So (being general here) there are r + (n−1) positions, and we want to choose r of them to have circles.

This is like saying "we have r + (n−1) pool balls and want to choose r of them". In other words it is now like the pool balls question, but with slightly changed numbers. And we can write it like this:

Interestingly, we can look at the arrows instead of the circles, and say "we have r + (n−1) positions and want to choose (n−1) of them to have arrows", and the answer is the same:

So, what about our example, what is the answer?

There are 35 ways of having 3 scoops from five flavors of icecream.

In Conclusion

Phew, that was a lot to absorb, so maybe you could read it again to be sure!

But knowing how these formulas work is only half the battle. Figuring out how to interpret a real world situation can be quite hard.

But at least you now know the 4 variations of "Order does/does not matter" and "Repeats are/are not allowed":

How to Solve Permutations and Combinations? (+FREE Worksheet!)

Learn how to solve mathematics word problems containing Permutations and Combinations using formulas.

How to Solve Permutations and Combinations? (+FREE Worksheet!)

Related Topics

Step by step guide to solve Permutations and Combinations

Permutations and Combinations – Example 1:

How many ways can first and second place be awarded to \(10\) people?

Since the order matters, we need to use the permutation formula where \(n\) is \(10\) and \(k\) is \(2\). Then: \(\frac{n!}{(n-k)!}=\frac{10!}{(10-2)!}=\frac{10!}{8!}=\frac{10×9×8!}{8!}\), remove \(8!\) from both sides of the fraction. Then: \(\frac{10×9×8!}{8!}=10×9=90\)

Permutations and Combinations – Example 2:

How many ways can we pick a team of \(3\) people from a group of \(8\)?

Since the order doesn’t matter, we need to use combination formula where \(n\) is \(8\) and \(r\) is \(3\). Then: \(\frac{n!}{r! (n-r)!}=\frac{8!}{3! (8-3)!}=\frac{8!}{3! (5)!}=\frac{8×7×6×5!}{3! (5)!}\), remove \(5!\) from both sides of the fraction. Then: \(\frac{8×7×6}{3×2×1}=\frac{336}{6}=56\)

Exercises for Solving Permutations and Combinations

Calculate the value of each..

Download Combinations and Permutations Worksheet

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MATH Worksheets 4 Kids

Combination Worksheets

Combination is a selection of distinct elements without specific orders. Combination worksheets are diligently prepared as per the state standards and proposed for high school students. Unearth the usage of combinations in real-world scenarios with this array of printable exercises, like listing out combinations, finding the number of combinations, evaluation, solving combination problems and more.We also have a huge collection of permutation worksheets for practice. Practice some of these worksheets for free!

Listing Out Combinations

Listing Out Combinations

Jot down all the possible combinations of the habitual elements like names, numbers, shapes, alphabets, colors and so on!

pdf 1

Number of Combinations

Write down the total number of possible combinations for a set of objects taken at a time. Each pdf worksheet consists of five problems.

exercise 1

Evaluate - Level 1

These printable high school worksheets include problems with simple expressions involving combinations. Students are required to simplify the expressions using the combination formula.

practice 1

Evaluate - Level 2

Level 2 worksheets include more integrative problems involving basic operations. Using the formula for C(n, r), evaluate the expressions.

worksheet 1

Solve - Level 1

Access this set of pdf combination worksheets to find the unknown value in the equations using the formula for C(n, r).

printable 1

Solve - Level 2

These printable Level 2 worksheets encompass challenging problems when compared to level 1 worksheets. Solve problems using the combination formula to know the unknown parameters.

Permutation and Combination - Mixed Review

Permutation and Combination - Mixed Review

These permutations and combinations worksheets consist of an array of exercises to identify and write permutations / combinations, twin-level of solving equations and evaluating expressions.

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Combinations – Example and Practice Problems

Combinations are used to count the number of different ways that certain groups can be chosen from a set if the order of the objects does not matter. This is different from permutations, where the order of the objects does matter.

Here, we will look at a brief summary of combinations along with their formula and the terminology used. In addition, we will see examples with answers to learn about the application of the combination formula.

examples of combinations

Relevant for …

Learning about combinations with solved examples.

See examples

Summary of combinations

Combinations – examples with answers, combinations – practice problems.

Combinations are selections of objects in a collection, in which the order of the selection does not matter. In combinations, we can select the objects in any order. For example, if we have  ab  and  ba , these selections are considered equal in combinations.

The formula to determine the number of possible organizations by selecting a few objects from a set without repetitions is expressed in the following way:

Recall that the factorial (denoted as “!”) is a product of all positive integers less than or equal to the number preceding the factorial. For example, $latex 3!=1 \times 2 \times 3 = 6$.

With the following examples, you can practice applying the combination formula. Each exercise has its respective solution to analyze the reasoning behind each answer.

Find the result of the combination $latex _{8}C_{6}$.

We use the formula of combinations $latex _{n}{{C}_{k}}=\frac{{n!}}{{\left( {n-k} \right)!k!}}$ and substitute $latex n=8$ and $latex k=6$:

$latex _{8}{{C}_{6}}=\frac{{8!}}{{\left( {8-6} \right)!6!}}$

$latex =\frac{{8!}}{{\left( {2} \right)!6!}}$

Now, we recognize that we can write to 8! like $latex 8 \times 7 \times 6!$ and we eliminate 6! both in the numerator and in the denominator:

$latex \frac{{8!}}{{\left( {2} \right)!6!}}=\frac{{8\times 7}}{2}$

$latex =4\times 7=28$

Find the result of the combination $latex _{9}C_{4}$.

We substitute $latex n=9$ and $latex k=4$ in the formula $latex _{n}{{C}_{k}}=\frac{{n!}}{{\left( {n-k} \right)!k!}}$:

$latex _{9}{{C}_{4}}=\frac{{9!}}{{\left( {9-4} \right)!4!}}$

$latex =\frac{{9!}}{{\left( {5} \right)!4!}}$

Now, we recognize that we can write to 9! like $latex 9\times 8\times 7\times 6\times 5!$ and we eliminate 5! both in the numerator and in the denominator:

$latex \frac{{9!}}{{\left( {5} \right)!4!}}=\frac{{9\times 8\times 7\times 6}}{4!}$

We rewrite 4! like $latex 4\times 3\times 2\times 1$ and simplify:

$latex \frac{{9\times 8\times 7\times 6}}{4\times 3\times 2\times 1}=126$

Find the combination $latex _{100}C_{100}$.

We substitute $latex n=100$ and $latex k=100$ in the formula $latex _{n}{{C}_{k}}=\frac{{n!}}{{\left( {n-k} \right)!k!}}$:

$latex _{100}{{C}_{100}}=\frac{{100!}}{{\left( {100-100} \right)!100!}}$

$latex =\frac{{100!}}{{\left( {1} \right)!100!}}$

We can easily eliminate 100! both denominator and numerator:

$latex \frac{{100!}}{{\left( {1} \right)!100!}}=1$

This result makes sense since there is only one possible way to select 100 objects from a set of 100 objects if the order does not matter.

How many ways are there to choose a team of 3 from a group of 10?

In this case, we choose 3 people so we have $latex k=3$. The whole group is $latex n=10$. Using this data in the formula $latex _{n}{{C}_{k}}=\frac{{n!}}{{\left( {n-k} \right)!k!}}$, we have:

$latex _{10}{{C}_{3}}=\frac{{10!}}{{\left( {10-3} \right)!3!}}$

$latex =\frac{{10!}}{{\left( {7} \right)!3!}}$

We can expand the 10! until you get 7! and we simplify this:

$latex \frac{10!}{(3)!3!}=\frac{10\times 9 \times 8 \times 7!}{(7)!3!}$

$latex =\frac{{10\times 9 \times 8}}{{3!}}$

$latex =\frac{{10\times 9 \times 8}}{{6}}$

$latex =120$

Suppose we have to select 5 new employees from a list of 10 applicants. In how many ways can this be done?

In this case, we have $latex n=10$ and $latex k=5$, therefore, we have:

$latex _{10}{{C}_{5}}=\frac{{10!}}{{\left( {10-5} \right)!5!}}$

$latex =\frac{{10!}}{{\left( {5} \right)!5!}}$

We can rewrite 10! until we get 5! and simplify:

$latex \frac{{10!}}{{\left( {5} \right)!5!}}=\frac{{10 \times 9\times 8 \times 7 \times 6 \times 5!}}{{\left( {5} \right)!5!}}$

$latex =\frac{{10 \times 9\times 8 \times 7 \times 6}}{{5!}}$

$latex =\frac{10 \times 9\times 8 \times 7 \times 6}{5 \times 4\times 3 \times 2 \times 1}$

$latex =252$

In a car dealership, there are 3 cars of a particular model that have to be transported to another dealership. If there are a total of 25 cars of this model, how many options are available to transport?

We recognize that we have $latex n=25$ y $latex k=3$ and we substitute these values in the formula $latex _{n}{{C}_{k}}=\frac{{n!}}{{\left( {n-k} \right)!k!}}$:

$latex _{25}{{C}_{3}}=\frac{{25!}}{{\left( {25-3} \right)!3!}}$

$latex =\frac{{25!}}{{\left( {22} \right)!3!}}$

We rewrite the factorial 25! until we get to 22!:

$latex \frac{{25!}}{{\left( {22} \right)!3!}}=\frac{{25 \times 24 \times 23 \times 22!}}{{\left( {22} \right)!3!}}$

Now, we simplify to 22! in the numerator and denominator:

$latex \frac{{25 \times 24 \times 23 \times 22!}}{{\left( {22} \right)!3!}}=\frac{{25 \times 24 \times 23}}{{3!}}$

$latex =25 \times 4 \times 23=2300$

Suppose we have an office of 5 women and 6 men and we have to select a committee of 4 people. In how many ways can we select 2 men and 2 women?

In this case, we have to find two different combinations and then multiply them. Therefore, we want to calculate $latex (_{5}{{C}_{2}})(_{6}{{C}_{2}})$. We can calculate these combinations separately:

$latex _{5}{{C}_{2}}=\frac{{5!}}{{\left( {5-2} \right)!2!}}$

$latex =\frac{{5!}}{{\left( {3} \right)!2!}}$

$latex =\frac{{5\times 4\times 3!}}{{\left( {3} \right)!2!}}$

$latex =\frac{{5\times 4}}{{2!}}=10$

$latex _{6}{{C}_{2}}=\frac{{6!}}{{\left( {6-2} \right)!2!}}$

$latex =\frac{{6!}}{{\left( {4} \right)!2!}}$

$latex =\frac{{6\times 5\times 4!}}{{\left( {4} \right)!2!}}$

$latex =\frac{{6\times 5}}{{2!}}=15$

Therefore, we have $latex (_{5}{{C}_{2}})(_{6}{{C}_{2}})=10\times 15=150$.

→ Combinations Calculator (nCr)

Put your knowledge of combinations into practice with the following problems. Solve the combinations and select an answer. Please check it to make sure you selected the correct one.

Find the combination $latex _{9}C_{5}$.

Choose an answer

Find the combinations $latex _{11}C_{9}$.

How many combinations are there if we take 2 objects from a set of 7, we want to choose a team of 6 out of a pool of 9. how many ways are there to accomplish this, there are 10 people in a meeting. if everyone shakes hands, how many handshakes are possible.

Interested in learning more about factorials, permutations, and combinations? Take a look at these pages:

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Combinations Calculator (nCr)

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The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Basically, it shows how many different possible subsets can be made from the larger set. For this calculator, the order of the items chosen in the subset does not matter.

Combinations Formula:

For n ≥ r ≥ 0.

The formula show us the number of ways a sample of “r” elements can be obtained from a larger set of “n” distinguishable objects where order does not matter and repetitions are not allowed. [1] "The number of ways of picking r unordered outcomes from n possibilities." [2]

Also referred to as r-combination or "n choose r" or the binomial coefficient .  In some resources the notation uses k instead of r so you may see these referred to as k-combination or "n choose k."

Combination Problem 1

Choose 2 Prizes from a Set of 6 Prizes

You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6. How many different combinations of 2 prizes could you possibly choose?

In this example, we are taking a subset of 2 prizes (r) from a larger set of 6 prizes (n). Looking at the formula, we must calculate “6 choose 2.”

C (6,2)= 6!/(2! * (6-2)!) = 6!/(2! * 4!) = 15 Possible Prize Combinations

The 15 potential combinations are {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, {5,6}

Combination Problem 2

Choose 3 Students from a Class of 25

A teacher is going to choose 3 students from her class to compete in the spelling bee. She wants to figure out how many unique teams of 3 can be created from her class of 25.

In this example, we are taking a subset of 3 students (r) from a larger set of 25 students (n). Looking at the formula, we must calculate “25 choose 3.”

C (25,3)= 25!/(3! * (25-3)!)= 2,300 Possible Teams

Combination Problem 3

Choose 4 Menu Items from a Menu of 18 Items

A restaurant asks some of its frequent customers to choose their favorite 4 items on the menu. If the menu has 18 items to choose from, how many different answers could the customers give?

Here we take a 4 item subset (r) from the larger 18 item menu (n). Therefore, we must simply find “18 choose 4.”

C (18,4)= 18!/(4! * (18-4)!)= 3,060 Possible Answers

Handshake Problem

In a group of n people, how many different handshakes are possible?

First, let's find the total handshakes that are possible. That is to say, if each person shook hands once with every other person in the group, what is the total number of handshakes that occur?

A way of considering this is that each person in the group will make a total of n-1 handshakes. Since there are n people, there would be n times (n-1) total handshakes. In other words, the total number of people multiplied by the number of handshakes that each can make will be the total handshakes. A group of 3 would make a total of 3(3-1) = 3 * 2 = 6. Each person registers 2 handshakes with the other 2 people in the group; 3 * 2.

Total Handshakes = n(n-1)

However, this includes each handshake twice (1 with 2, 2 with 1, 1 with 3, 3 with 1, 2 with 3 and 3 with 2) and since the orginal question wants to know how many different handshakes are possible we must divide by 2 to get the correct answer.

Total Different Handshakes = n(n-1)/2

Handshake Problem as a Combinations Problem

We can also solve this Handshake Problem as a combinations problem as C(n,2).

n (objects) = number of people in the group r (sample) = 2, the number of people involved in each different handshake

The order of the items chosen in the subset does not matter so for a group of 3 it will count 1 with 2, 1 with 3, and 2 with 3 but ignore 2 with 1, 3 with 1, and 3 with 2 because these last 3 are duplicates of the first 3 respectively.

expanding the factorials,

cancelling and simplifying,

which is the same as the equation above.

Sandwich Combinations Problem

This is a classic math problem and asks something like How many sandwich combinations are possible? and this is how it generally goes.

Calculate the possible sandwich combinations if you can choose one item from each of the four categories:

Often you will see the answer, without any reference to the combinations equation C(n,r), as the multiplication of the number possible options in each of the categories. In this case we calculate:

8 × 5 × 5 × 3 = 600 possible sandwich combinations

In terms of the combinations equation below, the number of possible options for each category is equal to the number of possible combinations for each category since we are only making 1 selection; for example C(8,1) = 8, C(5,1) = 5 and C(3,1) = 3 using the following equation:

C(n,r) = n! / ( r!(n - r)! )

We can use this combinations equation to calculate a more complex sandwich problem.

Sandwich Combinations Problem with Multiple Choices

Calculate the possible combinations if you can choose several items from each of the four categories:

Applying the combinations equation, where order does not matter and replacements are not allowed, we calculate the number of possible combinations in each of the categories. You can use the calculator above to prove that each of these is true.

Multiplying the possible combinations for each category we calculate:

8 × 10 × 10 × 8 = 6,400 possible sandwich combinations

How many possible combinations are there if your customers are allowed to choose options like the following that still stay within the limits of the total number of portions allowed:

In the previous calculation, replacements were not allowed; customers had to choose 3 different meats and 2 different cheeses. Now replacements are allowed, customers can choose any item more than once when they select their portions. For meats and cheeses this is now a combinations replacement or multichoose problem using the combinations with replacements equation:

C R (n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n - 1)!)

For meats, where the number of objects n = 5 and the number of choices r = 3, we can calculate either combinations replacement C R (5,3) = 35 or substitute terms and calculate combinations C(n+r-1, r) = C(5+3-1, 3) = C(7, 3) = 35 .

Calculating cheese choices in the same way, we now have the total number of possible options for each category at

and finally we multiply to find the total

8 × 35 × 15 × 8 = 33,600 possible sandwich combinations!

How many combinations are possible if customers are also allowed replacements when choosing toppings?

[1] Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 206, 2003.

For more information on combinations and binomial coefficients please see Wolfram MathWorld: Combination .

Cite this content, page or calculator as:

Furey, Edward " Combinations Calculator (nCr) " at https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators

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Combination Method for Solving Math Problems

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All these notations mean n choose r: C(n,r) = nCr = nCr = (nr) = n!r!(n-r)!. Just remember the formula: n!r!(n - r)!

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Combinations

In these lessons, we will learn the concept of combinations, the combination formula and solving problems involving combinations.

Related Pages Permutations Permutations and Combinations Counting Methods Factorial Lessons Probability

What Is Combination In Math?

An arrangement of objects in which the order is not important is called a combination. This is different from permutation where the order matters. For example, suppose we are arranging the letters A, B and C. In a permutation, the arrangement ABC and ACB are different. But, in a combination, the arrangements ABC and ACB are the same because the order is not important.

What Is The Combination Formula?

The number of combinations of n things taken r at a time is written as C( n , r ) .

The following diagram shows the formula for combination. Scroll down the page for more examples and solutions on how to use the combination formula.

Combination Formula

If you are not familiar with the n! (n factorial notation) then have a look the factorial lesson

How To Use The Combination Formula To Solve Word Problems?

Example: In how many ways can a coach choose three swimmers from among five swimmers?

Solution: There are 5 swimmers to be taken 3 at a time. Using the formula:

The coach can choose the swimmers in 10 ways.

Example: Six friends want to play enough games of chess to be sure every one plays everyone else. How many games will they have to play?

Solution: There are 6 players to be taken 2 at a time. Using the formula:

They will need to play 15 games.

Example: In a lottery, each ticket has 5 one-digit numbers 0-9 on it. a) You win if your ticket has the digits in any order. What are your changes of winning? b) You would win only if your ticket has the digits in the required order. What are your chances of winning?

Solution: There are 10 digits to be taken 5 at a time.

The chances of winning are 1 out of 252.

b) Since the order matters, we should use permutation instead of combination. P(10, 5) = 10 x 9 x 8 x 7 x 6 = 30240

The chances of winning are 1 out of 30240.

How To Evaluate Combinations As Well As Solve Counting Problems Using Combinations?

A combination is a grouping or subset of items. For a combination, the order does not matter.

How many committees of 3 can be formed from a group of 4 students? This is a combination and can be written as C(4,3) or 4 C 3 or \(\left( {\begin{array}{*{20}{c}}4\\3\end{array}} \right)\).

How To Solve Word Problems Involving Permutations And Combinations?

How To Solve Combination Problems That Involve Selecting Groups Based On Conditional Criteria?

Example: A bucket contains the following marbles: 4 red, 3 blue, 4 green, and 3 yellow making 14 total marbles. Each marble is labeled with a number so they can be distinguished.

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Combinations

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A combination is a way of choosing elements from a set in which order does not matter. A wide variety of counting problems can be cast in terms of the simple concept of combinations, therefore, this topic serves as a building block in solving a wide range of problems.

Introduction

Basic examples, intermediate examples, advanced examples, combinations with repetition, combinations with restriction, combinations - problem solving.

Consider the following example: Lisa has \(12\) different ornaments and she wants to give \(5\) ornaments to her mom as a birthday gift (the order of the gifts does not matter). How many ways can she do this?

We can think of Lisa giving her mom a first ornament, a second ornament, a third ornament, etc. This can be done in \( \frac{12!}{7!} \) ways. However, Lisa’s mom is receiving all five ornaments at once, so the order Lisa decides on the ornaments does not matter. There are \( 5! \) reorderings of the chosen ornaments, implying the total number of ways for Lisa to give her mom an unordered set of \(5\) ornaments is \( \frac{12!}{7!5!} \).

Notice that in the answer, the factorials in the denominator sum to the value in the numerator. This is not a coincidence. In general, the number of ways to pick \( k \) unordered elements from an \( n \) element set is \( \frac{n!}{k!(n-k)!} \). This is a binomial coefficient.

Proof of \(\displaystyle {n \choose k} = \frac{n!}{k!(n-k)!}:\)

Now suppose we want to choose \(k\) objects from \(n\) objects, then the number of combinations of \(k\) objects chosen from \(n\) objects is denoted by \(n \choose k\). Since \({_nP_k}=k!{n \choose k}\), it follows that

\[{n \choose k} = \frac{1}{k!}(_nP_k)= \frac{n!}{k!(n-k)!}.\]

How many ways are there to arrange 3 chocolate chip cookies and 10 raspberry cheesecake cookies into a row of 13 cookies? We can consider the situation as having 13 spots and filling them with 3 chocolate chip cookies and 10 raspberry cheesecake cookies. Then we just choose 3 spots for the chocolate chip cookies and let the other 10 spots have raspberry cheesecake cookies. The number of ways to do this job is \({13\choose3}=\frac{13\times12\times11}{3\times2\times1}=286.\) \(_\square\)
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? The number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 is \({7\choose3}\times{4\choose2} = 210.\) Therefore, the number of groups each containing 3 consonants and 2 vowels is \(210.\) Since each group contains 5 letters, which can be arranged amongst themselves in \(5!= 120\) ways, the required number of words is \(210\times120 = 25200.\ _\square\)
How many ways are there to select 3 males and 2 females out of 7 males and 5 females? The number of ways to select \(3\) males out of \(7\) is \({7 \choose 3} = \frac{7\times 6\times 5}{3 \times 2 \times 1}=35.\) Similarly, the number of ways to select \(2\) females out of \(5\) is \({5 \choose 2} = \frac{5\times 4}{2 \times 1}=10.\) Hence, by the rule of product, the answer is \(35 \times 10=350\) ways. \(_\square\)
There are \(9\) children. How many ways are there to group these \(9\) children into 2, 3, and 4? The number of ways to choose \(2\) children out of \(9\) is \({9\choose2}=\frac{9 \times 8}{2 \times 1}=36.\) The number of ways to choose \(3\) children out of \(9-2=7\) is \({7 \choose 3}=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35.\) Finally, the number of ways to choose \(4\) children out of \(7-3=4\) is \({4 \choose 4}=1.\) Hence, by the rule of product, the answer is \(36 \times 35 \times 1=1260\) ways. \(_\square\)
There are \(9\) distinct chairs. How many ways are there to group these chairs into 3 groups of 3? The number of ways to choose \(3\) chairs out of \(9\) is \({9\choose3}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84.\) The number of ways to choose \(3\) chairs out of \(9-3=6\) is \({6 \choose 3}=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20.\) Finally, the number of ways to choose \(3\) chairs out of \(6-3=3\) is \({3 \choose 3}=1.\) Now, since each of these three groups has an equal number of three chairs and the order of the three groups does not matter, by the rule of product our answer is \[\frac{84 \times 20 \times 1}{3 !}=280\] ways. \(_\square\)

At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

A combination is a way of choosing elements from a set in which order does not matter.

In general, the number of ways to pick \( k \) unordered elements from an \( n \) element set is \( \frac{n!}{k!(n-k)!} \). This is a binomial coefficient , denoted \( n \choose k \).

How many ordered non-negative integer solutions \( (a, b, c, d) \) are there to the equation \( a + b + c + d = 10 \)? To solve this problem, we use a technique called "stars and bars," which was popularized by William Feller. We create a bijection between the solutions to \( a + b + c + d = 10 \) and sequences of 13 digits, consisting of ten 1's and three 0's. Given a set of four integers whose sum is 10, we create a sequence that starts with \( a \) 1's, then has a 0, then has \( b \) 1's, then has a 0, then has \( c \) 1's, then has a 0, and then has \( d \) 1's. Conversely, given such a sequence, we can set \( a \) to be equal to the length of the initial string of 1's (before the first 0), set \( b \) equal to the length of the next string of 1's (between the first and second 0), set \( c \) equal to the length of the third string of 1's (between the second and third 0), and set \( d \) equal to the length of the fourth string of 1's. It is clear that such a procedure returns the starting set, and hence we have a bijection. Now, it remains to count the number of such sequences. We pick 3 positions for the 0's and the remaining positions are 1's. Hence, there are \( {13 \choose 3}= 286 \) such sequences. \(_\square\)
There are \(5\) shirts all of different colors, \(4\) pairs of pants all of different colors, and \(2\) pairs of shoes with different colors. In how many ways can Amy and Bunny be dressed up with a shirt, a pair of pants, and a pair of shoes each? We choose \(2\) shirts out of \(5\) for both Amy and Bunny to wear, so \({5\choose2}2!=20.\) We choose \(2\) pairs of pants out of \(4\) for them to wear, so \({4\choose2}2!=12.\) We choose \(2\) pairs of shoes out of \(2\) for them to wear, so \({2\choose2}2!=2.\) Therefore, by the rule of product, the answer is \(20 \times 12 \times 2=480\) ways. \(_\square\)
We are trying to divide 5 European countries and 5 African countries into 5 groups of 2 each. How many ways are there to do this under the restriction that at least one group must have only European countries? The number of ways to divide \(5+5=10\) countries into 5 groups of 2 each is as follows: \[\frac{{10\choose2} \times {8\choose2} \times {6\choose2} \times {4\choose2} \times {2\choose2}}{5 !} =\frac{ 45 \times 28 \times 15 \times 6 \times 1}{120}=945. \qquad (1)\] Since it is required that at least one group must have only European countries, we need to subtract from \((1)\) the number of possible groupings where all 5 groups have 1 European country and 1 African country each. This is equivalent to the number of ways to match each of the 5 European countries with one African country: \[5 ! = 5 \times 4 \times 3 \times 2 \times 1=120. \qquad (2)\] Therefore, taking \((1)-(2)\) gives our answer \(945-120=825.\) \(_\square\)

Find the number of rectangles in a \(10 \times 12\) chessboard.

Note: All squares are rectangles, but not all rectangles are squares.

There are two distinct boxes, 10 identical red balls, 10 identical yellow balls, and 10 identical blue balls. How many ways are there to sort the 30 balls into the two boxes so that each box has 15? Keeping in mind that the two boxes are distinct, let \(r, y\) and \(b\) be the numbers of red, yellow and blue balls in the first box, respectively. Then we first need to get the number of cases satisfying \(r+y+b=15,\) and then subtract the numbers of cases where \(r>10, y>10\) or \(b>10.\) Using stars and bars, the number of cases satisfying \(r+y+b=15\) is \({17\choose2}=136. \qquad (1)\) Now, the following gives the number of cases where \(10<r\le15:\) If \(r=11,\) then \(y+b=4,\) implying there are 5 such cases. If \(r=12,\) then \(y+b=3,\) implying there are 4 such cases. If \(r=13,\) then \(y+b=2,\) implying there are 3 such cases. If \(r=14,\) then \(y+b=1,\) implying there are 2 such cases. If \(r=15,\) then \(y+b=0,\) implying there is 1 such case. Hence, the number cases where \(10<r\le15\) is \(5+4+3+2+1=15.\) Since exactly the same logic applies for the cases where \(10<y\le15\) or \(10<b\le15,\) the total number of cases to subtract from \((1)\) is \(3\times 15=45. \qquad (2)\) Therefore, taking \((1)-(2)\) gives our answer \(136-45=91.\) \(_\square\)
PizzaHot makes 7 kinds of pizza, 3 of which are on sale everyday, 7 days a week. According to their policy, any two kinds of pizza that are on sale on a same day can never be on sale on the same day again during the rest of that calender week. Let \(X\) be the number of all the possible sale strategies during a calendar week. What is the remainder of \(X\) upon division by 1000? Let \(a, b, c, d, e, f, g\) be the 7 kinds of pizza. Then none of these 7 could be on sale for 4 days or more a week because each of the other 6 kinds would have been on sale on a same day in the first 3 days. In fact, since there are \(3\times 7=21\) pizzas on sale every week, each of the 7 kinds is on sale exactly 3 times a week. Now, without loss of generality, the number of ways to select 3 days to put \(a\) on sale is \({7\choose 3}.\) Then the number of ways to put each of the remaining 6 kinds on sale in those 3 days is \({6\choose 2}\times {4\choose 2}\times {2\choose 2} .\) The following table is one example of this operation, where \(a\) is on sale for all 3 days during the week, whereas each of the other 6 kinds is on sale only once: pizza Since we are done with \(a,\) we now put one \(b\) in each of 2 of the remaining 4 days, the number of ways of doing which is \({4 \choose 2}.\) Then, excluding \(c\) which was already on sale together with \(b\) on the first day, we put \(d, e, f, g\) in the 2 days where \(b\) was just put. However, since the combinations \((d, e)\) and \((f, g)\) were already used when dealing with \(a,\) the number of ways to put \(d, e, f, g\) in the two columns together with \(b\) is \({4 \choose 2}-2.\) Finally, we put one \(c\) in each of the remaining two columns and then fill the columns, the number of ways of doing which is 2. Hence, the number of all the possible sale strategies during a calendar week is \[{7\choose 3}\times {6\choose 2}\times {4\choose 2}\times {4\choose 2}\times\left({4 \choose 2}-2\right)\times 2=151200.\] Therefore, the remainder of \(151200\) upon division by 1000 is 200. \(_\square\)

Three squares are chosen at random on a chess board. Find the probability that they lie on any diagonal.

\(\) Note: A line connecting the three squares \((1,1), (2,3), (3,5)\) does not form a diagonal.

\(A\) and \(B\) are the only candidates who contest in an election. They secure \(11\) and \(7\) votes, respectively. In how many ways can this happen if it is known that \(A\) stayed ahead of \(B\) throughout the counting process of votes?

Main Article: Combinations with Repetition
You want to distribute 7 indistinguishable candies to 4 kids. If every kid must receive at least one candy, in how many ways can you do this? You first give one candy to each of the 4 kids to comply with the requirement that every kid must receive at least one candy. Then you are left with 3 candies to distribute to the 4 kids, which is equivalent to a problem of placing \(k=3\) indistinguishable balls into \(n=4\) labeled urns, which is known as balls and urns or stars and bars. Thus, our answer is \[\binom{n+k-1}{k} =\binom{n+k-1}{n-1}=\binom{3+4-1}{3}=20. \ _\square \]

Winston must choose 4 classes for his final semester of school. He must take at least 1 science class and at least 1 arts class. If his school offers 4 (distinct) science classes, 3 (distinct) arts classes and 3 other (distinct) classes, how many different choices for classes does he have?

\(\) Details and Assumptions:

How many six digit integers contain exactly four different digits?

Try more combinatorics problems.

Let \(x+y+z=m,\) where \(x, y, z\) are integers such that \(x\ge 1, y\ge 2, z\ge 3.\) If the number of ordered triples \((x, y, z)\) satisfying the equation is \(21,\) what is \(m?\) Let \(x-1=a, y-2=b, z-3=c,\) where \(a, b, c\) are non-negative because \(x\ge 1, y\ge 2, z\ge 3,\) then \[\begin{align} x+y+z&=m\\ (a+1)+(b+2)+(c+3)&=m\\ a+b+c&=m-6. \qquad (1) \end{align}\] Since the number of ordered non-negative integer triples \((a, b, c)\) satisfying \((1)\) is \(21,\) using the technique of stars and bars , we obtain \[\begin{align} \binom{3+(m-6)-1}{m-6} =\binom{m-4}{m-6}=\binom{m-4}{2}&=21\\ \frac{(m-4)(m-5)}{2!}&=21\\ m^2-9m+20&=42\\ m^2-9m-22&=0\\ (m+2)(m-11)&=0\\ m&=11 \end{align}\] because \(m>0.\ _\square\)

How many ways are there to select \(3\) numbers from the first \(20\) positive integers such that no 2 of the selected numbers are consecutive?

In the figure above with 9 squares, how many ways are there to select two squares which do not share an edge?

This problem is part of the set Countings.

Suppose a small country has \(15\) cities and \(70\) roads, where each road directly connects precisely \(2\) cities. What is the largest possible number of cities that are directly connected to every other city in the country?

A pawn is placed on the lower left corner square of a standard \(8\) by \(8\) chess board. A 'move' involves moving the pawn, where possible, either

Using these legitimate moves, the pawn is to be moved along a path from the lower left square to the upper right square.

How many such paths are there?

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To figure out this math problem, simply use the order of operations. First, solve the equation within the parentheses, then work with the exponents, then multiply and divide from left to right, and finally add and subtract from left to right.

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Combinations Formula With Solved Example Questions

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1.Consider an example problem where order does not matter and repetition is not allowed. In this kind of problem, you won't use the same item more than once

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Combinations (video lessons, examples and solutions)

Combinations Formula: For n r 0. Also referred to as r-combination or n choose r or the binomial coefficient. In some resources the notation uses k

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How to solve combination word problems

How To Solve Word Problems Involving Permutations And Combinations? A museum has 7 paintings by Picasso and wants to arrange 3 of them on the same wall. How

Combination Problems With Solutions

Clarify mathematic question

Combination Word Problems

Here are some carefully chosen combination word problems that will show you how to solve word problems involving combinations.

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Permutations and Combinations Problems

COMBINATION PROBLEMS WITH SOLUTIONS Problem 1 : Solution : Problem 2 : A box contains two white balls, three black balls and four red balls. Solution :.

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How to solve a combination problem

College algebra students dive into their studies How to solve a combination problem, and manipulate different types of functions.

Combinations Calculator (nCr)

Combinations formula with solved example questions, combination formula for counting and probability.

Combinations Formula: For n r 0. Also referred to as r-combination or n choose r or the binomial coefficient. In some resources the notation uses k

How to Calculate Combinations: 8 Steps (with Pictures)

Answer: Insert the given numbers into the combinations equation and solve. n is the number of items that are in the set (4 in this example) r is the number

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Combinations and permutations

To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial.Apr 8, 2022

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Combinations Worksheets

What Are Combination Probabilities? Probability is the measure of the chances and possibilities that will return a desired outcome. There are different types of probability and combination is one of its types. So, what are combination probabilities? When we are dealing with arrangement of objects and the order does not matter, then we can use combination probability. The result of this probability shows us the total number of ways objects can be arranged without any defined order. When there is an order to follow, a combination probability is called permutation. The formula to calculate combination probabilities include; C(n,r) = n!/((n-r)!r!) | n! means, n × (n-1) × (n-2) ,..., (n-(n-1)).

Basic Lesson

Introduces the fundamental Combinations. Provides a basic application. On the bookshelf, there are 4 math books, 5 history books and 3 geography books. If two books are selected randomly without replacement, what is the probability of getting: 1 math book and 1 history book?

Intermediate Lesson

This lesson focuses on using the Combinations with word problems. In the magician's hat, there are 4 rabbits, 5 handkerchiefs and 3 playing cards. If two items are selected randomly without replacement, what is the probability of getting 1 playing card and 1 handkerchief ?

Independent Practice 1

Students practice with 20 Combinations problems. The answers can be found below. A bag contains 4 red balls, 6 blue balls and 8 white balls. If two balls are selected one after one and first replaced; What is the probability of drawing...

Independent Practice 2

Another 20 Combinations problems. The answers can be found below.

Homework Worksheet

Reviews all skills in the unit. A great take home sheet. Also provides a practice problem.

10 problems that test Combinations skills.

Homework and Quiz Answer Key

Answers for the homework and quiz.

Answers for the lesson and practice sheets.

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IMAGES

  1. Solving a combination problem using a table

    solving combination math problems

  2. Combinations and Permutations Word Problems

    solving combination math problems

  3. In math, a combination explains how many ways you can arrange a group from a larger group. The

    solving combination math problems

  4. Problem Solving With Combinations

    solving combination math problems

  5. Expert Essay Writers

    solving combination math problems

  6. Solved: Use combinations to solve each problem. See Examples 1-...

    solving combination math problems

VIDEO

  1. Discrete Math || Combination Math

  2. 4.5 combination method

  3. Math 30-1: 11.3 Combinations

  4. Maths 95

  5. PROBLEMS IN PERMUTATION AND COMBINATION

  6. Lecture for 11 03 Combinations

COMMENTS

  1. Combinations (practice)

    Combinations Combinations CCSS.Math: HSS.CP.B.9 Google Classroom You might need: Calculator When a customer buys a family-sized meal at certain restaurant, they get to choose 3 3 side dishes from 9 9 options. Suppose a customer is going to choose 3 3 different side dishes. How many groups of 3 3 different side dishes are possible? Show Calculator

  2. Permutations & combinations (practice)

    Permutations & combinations. CCSS.Math: HSS.CP.B.9. Google Classroom. You need to put your reindeer, Prancer, Quentin, Rudy, and Jebediah, in a single-file line to pull your sleigh. However, Rudy and Prancer are best friends, so you have to put them next to each other, or they won't fly.

  3. Combination Problems With Solutions

    COMBINATION PROBLEMS WITH SOLUTIONS Problem 1 : Find the number of ways in which 4 letters can be selected from the word ACCOUNTANT. Solution : Find the number of ways in which 4 letters can be selected from the letters of the word ACCOUNTANT Problem 2 : A box contains two white balls, three black balls and four red balls.

  4. How to Calculate Combinations: 8 Steps (with Pictures)

    Solve the equation to find the number of combinations. You can do this either by hand or with a calculator. If you have a calculator available, find the factorial setting and use that to calculate the number of combinations. If you're using Google Calculator, click on the x! button each time after entering the necessary digits.

  5. Combinations and Permutations

    1. Permutations with Repetition These are the easiest to calculate. When a thing has n different types ... we have n choices each time! For example: choosing 3 of those things, the permutations are: n × n × n (n multiplied 3 times) More generally: choosing r of something that has n different types, the permutations are: n × n × ... (r times)

  6. How to Solve Permutations and Combinations? (+FREE Worksheet!)

    For a combination problem, use this formula: nCr = n! r!(n−r)! n C r = n! r! ( n − r)! Factorials are products, indicated by an exclamation mark. For example, 4! 4! Equals: 4×3× 2×1 4 × 3 × 2 × 1. Remember that 0! 0! is defined to be equal to 1 1. Permutations and Combinations - Example 1:

  7. Permutations and Combinations Problems

    This is a combination problem: combining 2 items out of 3 and is written as follows: n C r = n! / [ (n - r)! r! ] The number of combinations is equal to the number of permuations divided by r! to eliminates those counted more than once because the order is not important. Example 7: Calculate 3 C 2 5 C 5 Solution:

  8. Combination Worksheets

    Combination worksheets are diligently prepared as per the state standards and proposed for high school students. Unearth the usage of combinations in real-world scenarios with this array of printable exercises, like listing out combinations, finding the number of combinations, evaluation, solving combination problems and more.We also have a ...

  9. Combinations

    With the following examples, you can practice applying the combination formula. Each exercise has its respective solution to analyze the reasoning behind each answer. EXAMPLE 1 Find the result of the combination _ {8}C_ {6} 8C 6. Solution EXAMPLE 2 Find the result of the combination _ {9}C_ {4} 9C 4. Solution EXAMPLE 3

  10. Combination Method for Solving Math Problems

    The combination method for solving math problems involves adding equations together to get all but one of the variables out of the way, allowing us to solve for the remaining one. The process ...

  11. Combinations Calculator (nCr)

    Handshake Problem as a Combinations Problem We can also solve this Handshake Problem as a combinations problem as C (n,2). n (objects) = number of people in the group r (sample) = 2, the number of people involved in each different handshake

  12. Solving Word Problems Involving Combinations

    Solving Word Problems Involving Combinations Step 1: Identify the size of our set, call this n n. There may be more than one set! Step 2: Identify the size of the combination, call this m m...

  13. How to solve math problems using combination method

    The combination method for solving math problems involves adding equations together to get all but one of the variables out of the way, allowing us to solve Solve Now Why students love us Christopher Meyer. I'm in 7th grade and this app helps me out a lot, and it's always right too! :), this is the most and best useful app for math. ...

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    Type a math problem Solve algebra trigonometry Get step-by-step explanations See how to solve problems and show your work—plus get definitions for mathematical concepts Graph your math problems Instantly graph any equation to visualize your function and understand the relationship between variables Practice, practice, practice

  15. Finite Math Examples

    Finite Math Examples | Probability | Solving Combinations Finite Math Examples Step-by-Step Examples Finite Math Probability Find the Number of Possibilities 5C2 C 2 5 Evaluate 5C2 C 2 5 using the formula nCr = n! (n−r)!r! C r n = n! ( n - r)! r!. 5! (5−2)!2! 5! ( 5 - 2)! 2! Subtract 2 2 from 5 5. 5! (3)!2! 5! ( 3)! 2!

  16. Combinations (video lessons, examples and solutions)

    How To Solve Combination Problems That Involve Selecting Groups Based On Conditional Criteria? Example: A bucket contains the following marbles: 4 red, 3 blue, 4 green, and 3 yellow making 14 total marbles. Each marble is labeled with a number so they can be distinguished. How many sets/groups of 4 marbles are possible?

  17. Combinations

    A combination is a way of choosing elements from a set in which order does not matter. A wide variety of counting problems can be cast in terms of the simple concept of combinations, therefore, this topic serves as a building block in solving a wide range of problems. Contents Introduction Basic Examples Intermediate Examples Advanced Examples

  18. ALGEBRA How to Solve Math Problems on Combination

    Lesson on solving for combinations or fundamental counting principle. Combinations can be used to determine the number of ways or arrangements of a given pro...

  19. Problem solving involving permutation and combination

    Problem solving involving permutation and combination - THIRD QUARTER GRADE 10: SOLVING WORD PROBLEMS INVOLVING PERMUTATIONS AND COMBINATIONS GRADE 10. ... Solve math questions. If you're looking for help with math problems, you've come to the right place. Our experts are here to help you solve even the most difficult math questions.

  20. How to solve a combination math problem

    How to solve a combination math problem. In this blog post, we will provide you with a step-by-step guide on How to solve a combination math problem. Deal with math problems. Fast Delivery. Solve Now. Determine mathematic equations Enhance your scholarly performance Download full explanation ...

  21. How to solve combination word problems

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  23. How to solve a combination problem

    How to solve a combination problem - This site allow users to input a Math problem and receive step-by-step instructions on How to solve a combination problem. ... Math is a way of solving problems by using numbers and equations. 24/7 Live Specialist You can always count on our 24/7 customer support to be there for you when you need it ...

  24. Combinations Worksheets

    On the bookshelf, there are 4 math books, 5 history books and 3 geography books. If two books are selected randomly without replacement, what is the probability of getting: 1 math book and 1 history book? View worksheet. ... Students practice with 20 Combinations problems. The answers can be found below. A bag contains 4 red balls, 6 blue balls ...