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## Statistics and probability

## Permutations & combinations

- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text

## COMBINATION PROBLEMS WITH SOLUTIONS

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

Number of non black balls = 2 + 4 = 6

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

= ( 3 ⋅ 15) + ( 3 ⋅ 6) + (1 ⋅ 1)

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

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!

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)

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

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

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

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

Thus, the total number of ways is

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

## Calculating Combinations Without Repetition

- 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.
- For the example, you can calculate 10! with (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800. Find 4! with (4 * 3 * 2 * 1), which gives you 24. Find 6! with (6 * 5 * 4 * 3 * 2 * 1), which gives you 720.
- Then multiply the two numbers that add to the total of items together. In this example, you should have 24 * 720, so 17,280 will be your denominator.
- Divide the factorial of the total by the denominator, as described above: 3,628,800/17,280.
- In the example case, you'd do get 210. This means that there are 210 different ways to combine the books on a shelf, without repetition and where order doesn't matter.

## Calculating Combinations with Repetition

- If you have to solve by hand, keep in mind that for each factorial , you start with the main number given and then multiply it by the next smallest number, and so on until you get down to 0.
- For the example problem, your solution should be 11,628. There are 11,628 different ways you could order any 5 items from a selection of 15 items on a menu, where order doesn't matter and repetition is allowed.

## Community Q&A

## You Might Also Like

- ↑ https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php
- ↑ https://betterexplained.com/articles/easy-permutations-and-combinations/
- ↑ https://www.mathsisfun.com/combinatorics/combinations-permutations.html
- ↑ https://medium.com/i-math/combinations-permutations-fa7ac680f0ac
- ↑ https://www.quora.com/What-is-Combinations-with-repetition
- ↑ https://en.wikipedia.org/wiki/Combination
- ↑ https://mathbits.com/MathBits/TISection/Algebra1/Probability.htm

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## Combinations and Permutations

So, in Mathematics we use more precise language:

- When the order doesn't matter, it is a Combination .
- When the order does matter it is a Permutation .

A Permutation is an ordered Combination.

## Permutations

There are basically two types of permutation:

- Repetition is Allowed : such as the lock above. It could be "333".
- No Repetition : for example the first three people in a running race. You can't be first and second.

## 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:

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

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

## 2. Permutations without Repetition

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

## Example: what order could 16 pool balls be in?

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

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:

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

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

## 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):

- Repetition is Allowed : such as coins in your pocket (5,5,5,10,10)
- No Repetition : such as lottery numbers (2,14,15,27,30,33)

## 1. Combinations with Repetition

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

## 2. Combinations without Repetition

The easiest way to explain it is to:

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.

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":

## Example: Pool Balls (without order)

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

= 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 :

## Pascal's Triangle

OK, now we can tackle this one ...

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

- {c, c, c} (3 scoops of chocolate)
- {b, l, v} (one each of banana, lemon and vanilla)
- {b, v, v} (one of banana, two of vanilla)

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

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!

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

## Related Topics

- How to Interpret Histogram
- How to Interpret Pie Graphs
- How to Solve Probability Problems
- How to Find Mean, Median, Mode, and Range of the Given Data

## Step by step guide to solve Permutations and Combinations

- Permutations: The number of ways to choose a sample of \(k\) elements from a set of \(n\) distinct objects where order does matter, and replacements are not allowed. For a permutation problem, use this formula: \(\color{blue}{_{n}P_{k }= \frac{n!}{(n-k)!}}\)
- Combination: The number of ways to choose a sample of \(r\) elements from a set of \(n\) distinct objects where order does not matter, and replacements are not allowed. For a combination problem, use this formula: \(\color{blue}{_{ n}C_{r }= \frac{n!}{r! (n-r)!}}\)
- Factorials are products, indicated by an exclamation mark. For example, \(4!\) Equals: \(4×3×2×1\). Remember that \(0!\) is defined to be equal to \(1\).

## Permutations and Combinations – Example 1:

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

## Permutations and Combinations – Example 2:

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

## Exercises for Solving Permutations and Combinations

- \(\color{blue}{4!=}\)
- \(\color{blue}{4!×3!=}\)
- \(\color{blue}{5!=}\)
- \(\color{blue}{6!+3!=}\)
- There are \(7\) horses in a race. In how many different orders can the horses finish?
- In how many ways can \(6\) people be arranged in a row?

## Download Combinations and Permutations Worksheet

- \(\color{blue}{24}\)
- \(\color{blue}{144}\)
- \(\color{blue}{120}\)
- \(\color{blue}{726}\)
- \(\color{blue}{5,040}\)
- \(\color{blue}{720}\)

by: Reza about 3 years ago (category: Articles )

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## Combination Worksheets

Permutation and Combination - Mixed Review

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## Combinations – Example and Practice Problems

Learning about combinations with solved examples.

## Summary of combinations

Combinations – examples with answers, combinations – practice problems.

- n is the total number of elements in a set
- k is the number of selected objects
- ! is the factorial symbol

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

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

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

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

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

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

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

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

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 $latex _{100}C_{100}$.

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

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

We can easily eliminate 100! both denominator and numerator:

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

→ Combinations Calculator (nCr)

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

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

## Jefferson Huera Guzman

## Learn mathematics with our additional resources in different topics

Neurochispas is a website that offers various resources for learning Mathematics and Physics.

## INFORMATION

## Combinations Calculator (nCr)

## Calculator Use

## Combinations Formula:

## Combination Problem 1

Choose 2 Prizes from a Set of 6 Prizes

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

## Combination Problem 2

Choose 3 Students from a Class of 25

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

## Combination Problem 3

Choose 4 Menu Items from a Menu of 18 Items

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?

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).

which is the same as the equation above.

## Sandwich Combinations Problem

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

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

## Sandwich Combinations Problem with Multiple Choices

- 1 bread from 8 options is C(8,1) = 8
- 3 meats from 5 options C(5,3) = 10
- 2 cheeses from 5 options C(5,2) = 10
- 0 to 3 toppings from 3 options; we must calculate each possible number of choices from 0 to 3 and get C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8

Multiplying the possible combinations for each category we calculate:

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

- 2 portions of one meat and 1 portion of another?
- 3 portions of one meat only?
- 2 portions of one cheese only?

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

and finally we multiply to find the total

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

Cite this content, page or calculator as:

## How to solve math problems using combination method

Figure out mathematic question

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## Combination Method for Solving Math Problems

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## Combinations

## What Is Combination In Math?

## What Is The Combination Formula?

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

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.

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

They will need to play 15 games.

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

The chances of winning are 1 out of 252.

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.

- The soccer team has 20 players. There are always 11 players on the field. How many different groups of players can be on the field at any one time?
- A student need 8 more classes to complete her degree. If she met the prerequisites for all the courses, how many ways can she take 4 classes next semester?
- There are 4 men and 5 women in a small office. The customer wants a site visit from a group of 2 man and 2 women. How many different groups can be formed from the office?

## 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 many ways are there to do this?
- How many ways can you arrange the letters in the word LOLLIPOP?
- A person playing poker is dealt 5 cards. How many different hands could the player have been dealt?

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

- How many sets/groups of 4 marbles are possible?
- How many sets/groups of 4 are there such that each one is a different color?
- How many sets of 4 are there in which at least 2 are red?
- How many sets of 4 are there in which none are red, but at least one is green?

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## Combinations

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## Introduction

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

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

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. \(_\square\)

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

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. \(_\square\)

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

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 \[\binom{n+k-1}{k} =\binom{n+k-1}{n-1}=\binom{3+4-1}{3}=20. \ _\square \]

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 \(m>0.\ _\square\)

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## Easy Permutations and Combinations

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## How to solve a combination problem

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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

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.

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.

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.

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)

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:

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:

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 ...

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

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 ...

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

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...

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. ...

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

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!

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?

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

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...

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.

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 ...

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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 ...

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 ...