# Permutation word problems

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

Use the permutation formula shown below when the order is important.

Let nPr be the number of permutations of n objects arranged r at a time.

nPr = n(n - 1)(n - 2)(n - 3) ...

n is the first factor

Stop when there are r factors

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

• Use the permutation formula
• Use the multiplication principle and the permutation formula

Word problem #1

Eight cars enter a race. The three fastest cars will be given first, second, and third places.  How many arrangements of first, second, and third places are possible with eight cars?

Solution

Here, the order does matter since they are not just picking any 3 cars regardless of how fast they drive. They are picking the three fastest cars to give them first, second, third places.

Evaluate nPr with n = 8 and r = 3

8P3 = 8(8 - 1)(8 - 2)

8P3 = 8(7)(6) = 336

There are 336 possible arrangements of first, second, and third places.

Notice that once there are 3 factors, you stop!

Word problem #2

Tires on your cars should be rotated at regular intervals. How many ways can four tires be arranged?

Solution

Since all four tires are being rotated, you are using all the tires.

Evaluate nPr with n = 4 and r = 4

4P4 = 4(4 - 1)(4 - 2)(4 - 3)

4P4 = 4(3)(2)(1) = = (12)(2) = 24

The number of ways to arrange four tires on a car is 24.

Word problem #3

A baseball coach is going to pick 8 players from a baseball squad of 16 to take turn batting against the pitcher. How many batting orders are possible?

Solution

The total number of batting orders is the number of ways to arrange 8 players in order from a squad of 16.

Evaluate nPr with n = 16 and r = 8

16P8 = 16(16 - 1)(16 - 2)(16 - 3)(16 - 4)(16 - 5)(16 - 6)(16 - 7)

16P8 = 16(15)(14)(13)(12)(11)(10)(9)

16P8 = 518,918,400

Next time a baseball coach says that he had looked at all possible batting orders and picked the best ones, just say, "sure."

## More challenging permutation word problems

These permutation word problems will also show you how to use the multiplication principle to solve more complicated problems.

Word problem #4

A photographer is trying to take a picture of two men, three women, and four children. If the men, the women, and the children are always together, how many ways can the photographer arrange them?

Solution

Since the men, the women, and the children will stay together, we will have three groups. Below, see a picture of this situation. Then, the problem has the following four tasks:

Task 1: Find the number of ways the 2 men can be arranged (2P2)

Task 2: Find the number of ways the 3 women can be arranged (3P3)

Task 3: Find the number of ways the 4 children can be arranged (4P4)

Task 4: Find the number of ways the 3 groups can be arranged (3P3)

Then, use the fundamental counting principle shown below to find the total number of permutations of all 4 tasks.

Fundamental counting principle

If you have n choices for a first task and m choices for a second task, you have n × m choices for both tasks.

Therefore, evaluate 2P2 , 3P3 , 4P, and 3P3 and then multiply 2P2 , 3P3 , 4P4 , and 3P3 together.

2P2 = 2(2 - 1) = 2(1) = 2

3P3 = 3(3 - 1)(3 - 2) = 3(2)(1) = 6

4P4 = 4(4 - 1)(4 - 2)(4 - 3) = 4(3)(2)(1) = 24

3P3 = 6

2P2 × 3P3 × 4P4 × 3P3  = 2 × 6 × 24 × 6 = 1728

The photographer has 1728 ways to arrange these people. He better not make a big fuss about it!

Word problem #5

6 boys and 8 girls will have a presentation in class today. If the teacher is going to allow the girls to go first, how many different arrangement are there for the presentation?

Solution

If the girls present first, then the number of arrangement is 8P8

Then, when the boys present, the number of arrangements is 6P6

Using the multiplication principle, the total number of arrangement is 8P8 × 6P6

8P8 × 6P6 = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)(6 × 5 × 4 × 3 × 2 × 1)

8P8 × 6P6 = (40320)(720) = 29,030,400

## Permutation word problems with repetitions

Word problem #6

How many four- letter passwords can be made using the six letters a, b, c, d, e, and f?

With no repetitions, you can use the formula nPr = n(n - 1)(n - 2)(n - 3) ... and evaluate 6P4.

6P4 = 6(6 - 1)(6 - 2)(6 - 3) = 6 × 5 × 4 × 3 = 360

Notice that there are 6 choices for the first letter, 5 choices for the second letter, 4 choices for the third letter, and 3 choices for the fourth letter.

However, with repetitions, notice that there are 6 choices for the first letter, 6 choices for the second letter, 6 choices for the third letter, and 6 choices for the fourth letter.

6P4 with repetitions = 6 × 6 × 6 × 6 = 1296

Word problem #7

How many nine-letter passwords can be made using four a's, two b's, and three c's,?

This problem requires a special formula.

Let n be the number of items to be arranged.

Let n1 be items that are of one kind and are indistinguishable.

Let n2 be items that are of another kind and are indistinguishable.

Let nk be items that are of a kth kind and are indistinguishable.

Then you can use the formula you see below to find the number of distinguishable permutations:

n! / n1!n2!...nk!

In our problem above, there are nine letters to be arranged. So, let n  = 9

Four a's are indistinguishable. So let n1 = 4

Two b's are indistinguishable. So let n2 = 2

Three c's are indistinguishable. So let n3 = 3

Evaluate
9! / 4!×2!×3!

9×8×7×6×5×4! / 4!×2!×3!

9×8×7×6×5 / 2!×3!

9×8×7×6×5 / 2×3×2

15120 / 12
= 1260

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