Algebra II : Permutations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1 : Permutations

Find the Computing Permutation.

\displaystyle P (7, 5)

Possible Answers:

\displaystyle 5,100

\displaystyle 5,000

\displaystyle 5,400

\displaystyle 5,041

\displaystyle 2,520

Correct answer:

\displaystyle 2,520

Explanation:

\displaystyle P (7, 5)

\displaystyle \frac{7!}{(7 - 5)!} = \frac{7!}{2!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!} = \frac{5,040}{2}=2520 

Example Question #1 : How To Find The Number Of Integers Between Two Other Integers

An ice cream vendor sells five different flavors of ice cream. 

In how many ways can you choose three scoops of different ice cream flavors if order matters?

Possible Answers:

\displaystyle 60

\displaystyle 5^{3}

\displaystyle 35

\displaystyle 125

\displaystyle 3^{5}

Correct answer:

\displaystyle 60

Explanation:

There are five ways to choose the first scoop, then four ways to choose the second scoop, and finally three ways to choose the third scoop:

5 * 4 * 3 = 60

Example Question #1 : How To Find The Missing Number In A Set

There are 5 men and 4 women competing for an executive body consisting of :

  1. President
  2. Vice President
  3. Secretary
  4. Treasurer

It is required that 2 women and 2 men must be selected

How many ways the executive body can be formed?

Possible Answers:

\displaystyle 180

\displaystyle 1440

\displaystyle 60

\displaystyle 120

\displaystyle 240

Correct answer:

\displaystyle 1440

Explanation:

2 men can be selected:

2 women can be selected out of 4 women:

Finally, after the selection process, these men and women can fill the executive body in \displaystyle 4!=24 ways.

This gives us a total of \displaystyle 10\times 6\times 24 = 1440

Example Question #2 : Permutations

How many ways can a three committee board select the president, vice president and treasurer from a group of 15 people?

Possible Answers:

\displaystyle 3375

\displaystyle 2940

None of the above

\displaystyle 2700

\displaystyle 2730

Correct answer:

\displaystyle 2730

Explanation:

In this problem, order is important because once someone is chosen as a position they can not be chosen again, and once a position is filled, no one else can fill that in mind.

The presidential spot has a possibility of 15 choices, then 14 choices for vice president and 13 for the treasurer.

So:

\displaystyle 15*14*13=2730

Example Question #2 : Permutations

How many ways can you re-arrange the letters of the word JUBILEE?

Possible Answers:

\displaystyle 5,040

\displaystyle 5,460

\displaystyle 10,920

\displaystyle 2,520

Correct answer:

\displaystyle 2,520

Explanation:

There are 7 letters in the word jubilee, so initially we can calculate that there are \displaystyle 7! = 5,040 ways to re-arrange those letters. However, The letter e appears twice, so we're double counting. Divide by 2 factorial (2) to get \displaystyle 2,520.

Example Question #3 : Permutations

How many ways can you re-arrange the letters of the word BANANA?

Possible Answers:

\displaystyle 18

\displaystyle 20

\displaystyle 40

\displaystyle 60

\displaystyle 720

Correct answer:

\displaystyle 60

Explanation:

At first, it makes sense that there are \displaystyle 6! = 720 ways to re-arrange these letters. However, the letter A appears 3 times and the letter N appears twice, so divide first by 3 factorial and then 2 factorial:

\displaystyle \frac{720}{3! \cdot 2! } = \frac{720}{6 \cdot 2 } = \frac{720 } { 12} = 60

Example Question #42 : Algebra Ii

In a class of 24 students, how many distinct groups of 4 can be formed?

Possible Answers:

\displaystyle 10,626

\displaystyle 1,153

\displaystyle 720

\displaystyle 6

Correct answer:

\displaystyle 10,626

Explanation:

To solve, evaluate

Example Question #42 : Algebra Ii

13 rubber ducks are competing in a race. How many different arrangements of first, second, and third place are possible?

Possible Answers:

\displaystyle 1,716

\displaystyle 6

\displaystyle 1,037,836,800

\displaystyle 286

\displaystyle 3,628,800

Correct answer:

\displaystyle 1,716

Explanation:

There are 3 winners out of the total set of 13. That means we're calculating \displaystyle 13P3 = \frac{13! }{(13-3)!} = \frac{13!}{10!} = 1,716

Example Question #3 : Permutations

7 students try out for the roles of Starsky and Hutch in a new school production. How many different ways can these roles be cast?

Possible Answers:

\displaystyle 21

\displaystyle 120

\displaystyle 42

\displaystyle 5

\displaystyle 2,520

Correct answer:

\displaystyle 42

Explanation:

There are 7 potential actors and 2 different roles to fill. This would be calculated as \displaystyle 7! divided by \displaystyle (7-2)! = 5! , or 42

Example Question #4 : Permutations

How many different 4 letter words can be made out of the letters A, B, C, D, and E?

Possible Answers:

\displaystyle 20

\displaystyle 5

\displaystyle 60

\displaystyle 120

\displaystyle 480

Correct answer:

\displaystyle 120

Explanation:

Since order matters, use the permutation formula:

\displaystyle P(n,r)=\frac{n!}{(n-r)!}

There are 5 letters to choose from (n), and you pick 4 of them (r).

\displaystyle P(5,4)=\frac{5!}{(5-4)!}=\frac{5*4*3*2*1}{1}=120

So there are 120 possible words that can be formed.

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