Combination word problems with solutions

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

Use the combination formula shown below when the order does not matter

The combination word problems will show you how to do the followings:

• Use the combination formula

• Use the multiplication principle and the combination formula

• Use the addition principle and the combination formula

• Use the multiplication principle, addition principle, and combination formula

₁₈C₁₁ = 31824

There are 31824 different eleven-person students that can be chosen from a group of 18 students.

There are 4 different ways you can choose 3 animals from a list of 4.

Evaluate _nC_r with n = 10 and r = 4

There are 210 ways the teacher can pick students

More challenging combination word problems

These combination word problems will also show you how to use the multiplication principle and the addition principle.

Solution

This problem has the following two tasks:

Task 1: choose 3 males from 20 male employees

Task 2: choose 2 females from 30 female employees

We need to use the fundamental counting principle, also called the multiplication principle, since we have more than 1 task.  

Therefore, evaluate ₂₀C₃ and ₃₀C₂ and then multiply ₂₀C₃ by ₃₀C₂

₂₀C₃ × ₃₀C₂ = 1140 × 435 = 495900

The number of ways the committee can be chosen is 495900

The company can choose 2 people, 1 person, or none.

Notice that this time we need to use the addition principle as opposed to using the multiplication principle.

What is the difference? The key difference here is that the company will choose either 2, 1, or none. The company will not choose 2 people and 1 person at the same time. This does not make sense!

Therefore you need to evaluate ₈C₂, ₈C₁, and ₈C₀ and then add ₈C₂, ₈C₁, and ₈C₀ together.

The company has 39 ways to choose two or fewer candidates. 

Solution

The expression as many as makes the problem quite complex now since we now have all the following cases to consider.

Choose 3 males, 2 males, 1 male, or 0 male

Choose 2 females, 1 female, or 0 female.

Here is a complete list of all the different cases.

• 3 males and 2 females
• 3 males and 1 female
• 3 males and 0 female
• 2 males and 2 females
• 2 males and 1 female
• 2 males and 0 female
• 1 male and 2 females
• 1 male and 1 female
• 1 male and 0 female
• 0 male and 2 females
• 0 male and 1 female
• 0 male and 0 female

3 males and 2 females: ₂₀C₃ × ₃₀C₂ = 1140 × 435 = 495900 (done in problem #4)

3 males and 1 female: ₂₀C₃ × ₃₀C₁ = 1140 × 30 = 34200 

3 males and 0 female: ₂₀C₃ × ₃₀C₀ = 1140 × 1 = 1140 

2 males and 2 females: ₂₀C₂ × ₃₀C₂ =  190 × 435 = 82650

2 males and 1 female: ₂₀C₂ × ₃₀C₁ = 190 × 30 = 5700

2 males and 0 female: ₂₀C₂ × ₃₀C₀ = 190 × 1 = 190 

1 male and 2 females: ₂₀C₁ × ₃₀C₂ = 20 × 435 = 8700 

1 male and 1 female: ₂₀C₁ × ₃₀C₁ = 20 × 30 = 600 

1 male and 0 female: ₂₀C₁ × ₃₀C₀ = 20 × 1 = 20 

0 male and 2 females: ₂₀C₀ × ₃₀C₂ = 1 × 435 = 435

0 male and 1 female: ₂₀C₀ × ₃₀C₁ = 1 × 30 = 30

0 male and 0 female: ₂₀C₀ × ₃₀C₀ = 1 × 1 = 1

Add everything:

495900 + 34200 + 1140 + 82650 + 5700 + 190 + 8700 + 600 + 20 + 435 + 30 + 1 = 629566.

The number of ways to choose the committee is 629566

Problem 1 :

Find the total number of subsets of a set with

[Hint: ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ]

(i) 4 elements (ii) 5 elements (iii) n elements

Solution :

(i)   4 elements 

ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ

here n = 4

  =  ⁴C₀ + ⁴C₁ + ⁴C₂ + ⁴C₃  + ⁴C₄

  = 2⁴

  =  16

(ii)   5 elements 

ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ

here n = 5

    = 2⁵

  =  32

(iii)   n elements 

ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ

here n = n

    = 2ⁿ elements 

Problem 2 :

A trust has 25 members.

(i) How many ways 3 officers can be selected?

(ii) In how many ways can a President, Vice President and a Secretary be selected?

Solution :

(i)  Out of 25 members only 3 officers can be selected.

number of ways of selecting 3 officers  =  ²⁵C₃

  =  25!/22! 3!

  =  (25 ⋅ 24 ⋅ 23)/6

  =  2300

(ii) To select a president, we have 9 options 

to select vice president, we have 8 options

to select secretary, we have 7 options

total number of ways  =  9 ⋅ 8 ⋅ 7  =  504

Hence the answer is 504.

Problem 3 :

How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?

Solution :

Out of 10 members, we have to select a chair person. So we have 10 options to select a chair person. 9 options to select a secretary.

After selecting a chair person and secretary, we have 8 member. Out of 8, we have to select 4 persons.

Hence the answer is (10 ⋅ 9) ⁸C₄

Problem 4 :

How many different selections of 5 books can be made from 12 different books if,

(i) Two particular books are always selected?

(ii) Two particular books are never selected?

Solution :

(i)  Since two particular books are always selected, we may select remaining 3 books out of 10 books.

¹⁰C₃ =  10!/(7! 3!)  =  (10 ⋅ 9 ⋅ 8)/( 3 ⋅ 2) 

  =  120

Hence the answer is 120. 

(ii)  Since two particular books are never selected, we may select 5 books out of 10 books.

¹⁰C₅  =  10!/(5! 5!)  =  (10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6)/(5 ⋅ 4 ⋅ 3 ⋅ 2) 

  =  252

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Exercise 10

How many groups can be made from the word «house» if each group consists of 3 alphabets?

Exercise 11

Sarah has 8 colored pencils that are all unique. She wants to pick three colored pencils from her collection and give them to her younger sister. How many different combinations of colored pencils can Sarah make from 8 pencils?

Exercise 12

Alice has 6 chocolates. All of the chocolates are of different flavors. She wants to give two of her chocolates to her friend. How many different combinations of chocolates can Alice make from six chocolates?

Solution of exercise 1

How many different combinations of management can there be to fill the positions of president, vice-president and treasurer of a football club knowing that there are 12 eligible candidates?

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 2

How many different ways can the letters in the word «micro» be arranged if it always has to start with a vowel?

The words will begin with i or o followed by the remaining 4 letters taken from 4 by 4.

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 3

How many combinations can the seven colors of the rainbow be arranged into groups of three colors each?

The order of the elements does not matter.

The elements cannot be repeated.

Solution of exercise 4

How many different five-digit numbers can be formed with only odd numbered digits? How many of these numbers are greater than 70,000?

The order of the elements does matter.

The elements cannot be repeated.

n = 5      k = 5

The odd numbers greater than 70,000 have to begin with 7 or 9. Therefore:

Solution of exercise 5

How many games will take place in a league consisting of four teams? (Each team plays each other twice, once at each teams respective «home» location)

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 6

10 people exchange greetings at a business meeting. How many greetings are exchanged if everyone greets each other once?

The order of the elements does not matter.

The elements cannot be repeated.

Solution of exercise 7

How many five-digit numbers can be formed with the digits 1, 2 and 3? How many of those numbers are even?

The order of the elements does matter.

The elements are repeated.

If the number is even it can only end in 2.

Solution of exercise 8

How many lottery tickets must be purchased to complete all possible combinations of six numbers, each with a possibility of being from 1 to 49?

The order of the elements does not matter.

The elements cannot be repeated.

Solution of exercise 9

How many ways can 11 players be positioned on a soccer team considering that the goalie cannot hold another position other than in goal?

Therefore, there are 10 players who can occupy 10 different positions.

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 10

The word house has 5 alphabets. If each new word should have 3 alphabets, then we should use the following formula:

, where

Substitute the values in this example in the above formula:

Solution of exercise 11

Number of pencils Sarah have = 8

Number of pencils she wants to give to her younger sister = 3

We will use the binomial coefficients formula to determine the number of combinations:

, where

After substitution we will get the number of combinations:

Hence, Sarah can make 56 combinations of 8 colored pencils given the fact that she can choose 3 at a time.

Solution of exercise 12

Number of chocolates Alice have = 6

Number of chocolates she wants to give to her friend= 2

We will use the binomial coefficients formula to determine the number of combinations:

, where

After substitution we will get the number of combinations:

Hence, Alice can make 15 combinations of 6 chocolates given the fact that she can choose 2.

In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions.

Definition

Permutations are the different ways in which a collection of items can be arranged.

For example:

The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC.

Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.

The number of ways in which n things can be arranged, taken all at a time, ⁿP_n = n!, called ‘n factorial.’

Factorial Formula

Factorial of a number n is defined as the product of all the numbers from n to 1.

For example, the factorial of 5, 5! = 5*4*3*2*1 = 120.

Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways.

Number of permutations of n things, taken r at a time, denoted by:

ⁿP_r = n! / (n-r)!

For example:

The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways.

Important Permutation Formulas

1! = 1

Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.

Solution:

‘CHAIR’ contains 5 letters.

Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120.

 
Problem 2: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.

Solution:

The word ‘INDIA’ contains 5 letters and ‘I’ comes twice.

When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter.

Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60.

 
Problem 3: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?

Solution:

The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice.

Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.

 
Problem 4: How many different words can be formed with the letters of the word ‘SUPER’ such that the vowels always come together?

Solution:

The word ‘SUPER’ contains 5 letters.

In order to find the number of permutations that can be formed where the two vowels U and E come together.

In these cases, we group the letters that should come together and consider that group as one letter.

So, the letters are S,P,R, (UE). Now the number of words are 4.

Therefore, the number of ways in which 4 letters can be arranged is 4!

In U and E, the number of ways in which U and E can be arranged is 2!

Hence, the total number of ways in which the letters of the ‘SUPER’ can be arranged such that vowels are always together are 4! * 2! = 48 ways.

 
Problem 5: Find the number of different words that can be formed with the letters of the word ‘BUTTER’ so that the vowels are always together.

Solution:

The word ‘BUTTER’ contains 6 letters.

The letters U and E should always come together. So the letters are B, T, T, R, (UE).

Number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice).

Number of ways in which U and E can be arranged = 2! = 2 ways

Solution:

The word ‘REMAINS’ has 7 letters.

There are 4 consonants and 3 vowels in it.

Writing in the following way makes it easier to solve these type of questions.

(1) (2) (3) (4) (5) (6) (7)

No. of ways 3 vowels can occur in 4 different places = ⁴P₃ = 24 ways.

After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = ⁴P₄ = 4! = 24 ways.

Therefore, total number of permutations possible = 24*24 = 576 ways.

Combinations

Definition

The different selections possible from a collection of items are called combinations.

For example:

The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA.

It does not matter whether we select A after B or B after A. The order of selection is not important in combinations.

To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by ⁿC_r is

ⁿC_r = n! / [r! * (n-r)!]

For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are

³C₂ = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)

Important Combination formulas

ⁿC_n = 1

ⁿC₀ = 1

ⁿC₁ = n

ⁿC_r = ⁿC_(n-r)

The number of selections possible with A, B, C, taken all at a time is ³C₃ = 1 (i.e. ABC)

Solved examples of Combination

Let us take a look at some examples to understand how Combinations work:

 
Problem 1: In how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?

Solution:

No. of ways 1 man can be selected from a group of 3 men = ³C₁ = 3! / 1!*(3-1)! = 3 ways.

No. of ways 3 women can be selected from a group of 4 women = ⁴C₃ = 4! / (3!*1!) = 4 ways.

Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.

Solution:

Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be

3 B and 2 R

4 B and 1 R and

5 B and 0 R balls.

Therefore, our solution expression looks like this.
⁵C₃ * ³C₂ + ⁵C₄ * ³C₁ + ⁵C₅ * ³C₀ = 46 ways .

 
Problem 3: How many 4 digit numbers that are divisible by 10 can be formed from the numbers 3, 5, 7, 8, 9, 0 such that no number repeats?

Solution:

If a number is divisible by 10, its units place should contain a 0.
_ _ _ 0

After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits.

Selecting one digit out of 5 digits can be done in ⁵C₁ = 5 ways.

After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in ⁴C₁ = 4 ways.

After filling the hundreds place, the thousands place can be filled in ³C₁ = 3 ways.

Therefore, the total combinations possible = 5*4*3 = 60.

Permutations and Combinations Quiz

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Problem 1: Click here

Answer 1: Click here

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Problem 2: Click here

Answer 2: Click here

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Problem 3: Click here

Answer 3: Click here

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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.

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.

a) Using the formula:

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 ₄C₃
or (left( {begin{array}{*{20}{c}}4\3end{array}} right)).

Examples:

1. 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?
2. 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?
3. 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?

• Show Video Lesson

How To Solve Word Problems Involving Permutations And Combinations?

Examples:

1. 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?
2. How many ways can you arrange the letters in the word LOLLIPOP?
3. A person playing poker is dealt 5 cards. How many different hands could the player have been
   dealt?

• Show Video Lesson

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.

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

• Show Video Lesson

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