Algebra II - Permutations

Evaluate: $_{10}P_7$

Explanation

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:

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

50%

25%

100%

Cannot be determined

Explanation

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%.

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

Explanation

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

Evaluate:

Explanation

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:

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

Explanation

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.

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

Explanation

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:

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?