Permutations - Algebra II

Card 0 of 19

Question

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?

Answer

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

Compare your answer with the correct one above

Question

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

President

Vice President

Secretary

Treasurer

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

How many ways the executive body can be formed?

Answer

2 men can be selected:

$C\binom{5}{2 } = 10$

2 women can be selected out of 4 women:

$C\binom{4}{2} = 6$

Finally, after the selection process, these men and women can fill the executive body in ways.

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

Compare your answer with the correct one above

Question

Find the Computing Permutation.

Answer

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

Compare your answer with the correct one above

Question

There are runners in a race. How many different arrangements are there for $1^{st}$ , $2^{nd}$ , and $3^{rd}$ place?

Answer

This is a permutation of 10 objects (runners) taken 3 at a time, with no replacements.

$_{n}P_{k}=\frac{n!}{(n-k)!}$

$_{10}P_{3}=\frac{10!}{(10-3)!}=\frac{10!}{7!}=720$

Another way to look at this would be there are 10 runners competing for 1st place, 9 runners competing for 2nd place, and 8 runners competing for 3rd place.

Compare your answer with the correct one above

Question

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

Answer

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:

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

At first, it makes sense that there are 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:

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

Compare your answer with the correct one above

Question

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

Answer

To solve, evaluate $\binom{24}{4} = \frac{24! }{4! \cdot 20!} = 10,626$

Compare your answer with the correct one above

Question

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?

Answer

There are 7 potential actors and 2 different roles to fill. This would be calculated as divided by , or 42

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

Since order matters, use the permutation formula:

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

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

$P(5,4)=\frac{5!}{(5-4)!}=\frac{54321}{1}=120$

So there are 120 possible words that can be formed.

Compare your answer with the correct one above

Question

The soccer team awards gold, silver, and bronze trophies for the top three goal scorers over the season. If the soccer team has 11 players, how many different ways could the gold, silver, and bronze trophies be awarded?

Answer

Because the order is important, this is a permutation. There are 11 players to fill 3 spots, so .

Compare your answer with the correct one above

Question

How many possible ways can the letters A and B be assigned to nine people?

Answer

Write the formula for permutation.

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

Evaluate .

$\frac{9!}{(9-2)!} =\frac{9!}{7!} =\frac{9\times8\times7\times6\times5\times 4\times3\times2\times1}{7\times6\times5\times 4\times3\times2\times1}$

Cancel all the common terms in the numerator and denominator.

$9\times 8 = 72$

The answer is:

Compare your answer with the correct one above

Question

Evaluate: $_{10}P_7$

Answer

Write the permutation formula.

$_{n}P_r = \frac{n!}{(n-r)!}$

Substitute the values in the given problem.

$_{10}P_7=\frac{10!}{(10-7)!} = \frac{10\times 9\times8\times7\times6\times5\times4\times3\times2\times1}{7\times6\times5\times4\times3\times2\times1}$

Cancel out the terms in the numerator and denominator.

The remaining terms in the numerator are:

$10 \times 9 \times 8= 720$

The answer:

Compare your answer with the correct one above

Question

Evaluate:

Answer

Write the permutation formula.

$_nP_r =\frac{n!}{(n-r)!}$

Substitute the values of the variables.

$_6P_4 =\frac{6!}{(6-4)!}$

Simplify the factorials.

$\frac{6!}{2!} = \frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1}$

Cancel the common terms.

$6\times 5\times 4\times 3 =360$

The answer is:

Compare your answer with the correct one above

Question

Determine the value of:

Answer

Write the permutation formula.

$_nP_k = \frac{n!}{(n-k)!}$

Substitute the values of into the formula.

$_7P_4 = \frac{7!}{(7-4)!} = \frac{7!}{3!}$

Expand the terms.

$7!= 7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$

$3!=3\cdot2\cdot1$

$\frac{7!}{3!}= \frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}$

Cancel out the terms in the denominator and denominator.

The remaining terms are: $7\cdot6\cdot5\cdot4$

Multiply these numbers.

The answer is:

Compare your answer with the correct one above

Question

Solve the permutation:

Answer

Write the formula for permutation.

$_nP_r =\frac{n!}{(n-r)!}$

Substitute the values and expand the factorials.

$_9P_5=\frac{9!}{(9-5)!} = \frac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}$

Reduce the terms.

$9\times 8\times 7\times 6\times 5= 15120$

The answer is:

Compare your answer with the correct one above

Question

Suppose a locksmith designs a three digit lock with numbers ranging only from zero to five. What is the number of permutations possible?

Answer

The formula to determine the permutation is:

Notice that for each number, there are 6 different choices since we have numbers ranging from zero to five, not five.

$[0,1,2,3,4,5]\rightarrow n=6$

For a three digit lock, .

Substitute the terms into the formula.

The answer is:

Compare your answer with the correct one above

Question

What is the probability of flipping exactly one heads and exactly one tails in two coin tosses?

Answer

Each of the two coin flips represents one event. The probability of obtaining exactly one heads and exactly one tails can be modeled as follows:

Event 1 * Event 2

The key to understanding this problem is to recognize that either heads followed by tails, or tails followed by heads, would satisfy the specific overall outcome asked for in the problem. Since Event 1 will produce either a heads or a tails, we have a 100% chance of obtaining an outcome from Event 1 that will satisfy one of the two requirements of the specific overall outcome (one heads and one tails). Event 2 will then have a 50% chance of producing an outcome that is the opposite, rather than the same, as the outcome of Event 1. We can therefore calculate the probability of our specific overall outcome as follows:

100% * 50%

1 * 0.5

0.5 = 50%

Therefore, the probability of obtaining one heads and one tails from two coin tosses is 50%.

Compare your answer with the correct one above

Tap the card to reveal the answer