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Pearson Geometry Common Core, 2011

PG

Pearson Geometry Common Core, 2011 View details

Mid-Chapter Quiz

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Exercise 14 Page 843

To evaluate the given permutation, we will use the corresponding formula.

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

We are given that

n = 15

and

r = 5 .

Let's substitute these values into the formula.

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

$n = 15$ , $r = 5$

_{15} P_{5} = \frac{1 5 !}{( 1 5 - 5 ) !}

Subtract term

_{15} P_{5} = \frac{1 5 !}{1 0 !}

Write as a product

_{15} P_{5} = \frac{1 5 \cdot 1 4 \cdot 1 3 \cdot 1 2 \cdot 1 1 \cdot 1 0 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{1 0 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}

CancelCommonFac

Cancel out common factors

_{15} P_{5} = \frac{1 5 \cdot 1 4 \cdot 1 3 \cdot 1 2 \cdot 1 1 \cdot 1 0 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{1 0 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}

Simplify quotient

_{15} P_{5} = \frac{1 5 \cdot 1 4 \cdot 1 3 \cdot 1 2 \cdot 1 1}{1}

$\frac{a}{1} = a$

_{15} P_{5} = 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11

Multiply

_{15} P_{5} = 360360

There are

360360

permutations.

Alternative Solution

Using the calculator

We can evaluate the number of permutations $_{15} C_{5}$ using the graphic calculator. To do so, we have to start by entering the number of items, which is equal to $15 .$

Next, we push $M A T H$ and then scroll right until we reach PRB. Then, scroll down to the second row and push $E N T E R .$

window of a TI83 graphing calculator

window of a TI83 graphing calculator

Finally, by entering the number of items to be chosen and hitting $E N T E R,$ we can calculate the number of permutations.