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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 15 Page 843

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

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

We are given that

n = 4

and

r = 3 .

Let's substitute these values into the formula.

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

$n = 4$ , $r = 3$

_{4} C_{3} = \frac{4 !}{( 4 - 3 ) ! 3 !}

Subtract term

_{4} C_{3} = \frac{4 !}{3 ! 1 !}

Write as a product

_{4} C_{3} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{( 3 \cdot 2 \cdot 1 ) ( 1 )}

CancelCommonFac

Cancel out common factors

_{4} C_{3} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{( 3 \cdot 2 \cdot 1 ) ( 1 )}

Simplify quotient

_{4} C_{3} = \frac{4}{1}

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

_{4} C_{3} = 4

There are

4

combinations.

Alternative Solution

Using the calculator

We can evaluate the number of combinations $_{4} C_{3}$ using the graphic calculator. To do so, we have to start by entering the number of items, which is equal to $4 .$

Next, we push $M A T H$ and then scroll right until we reach PRB. Then, scroll down to the third 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 combinations.