Exercise 20 Page 506

We are asked to find in how many ways a $3 -$ person dart team can be chosen from $12$ people. To answer this, we must determine if this is a permutation or a combination. For a more detailed explanation of the difference between the concepts, see the end of the exercise.

A permutation is the number of ways you can choose and arrange a set of items, where the order that an item is chosen matters.
A combination is the number of ways you can choose and arrange a set of items where the order an item is chosen does not matter. What this means is that we disregard repeated elements.

When choosing the three-person dart team, the order in which the people are selected does not matter. Think about it — if you are selected first, second, or third does not matter, as long as you make the group. Therefore, we want to know the number of combinations. To calculate this we use the following formula.

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

Since we have twelve people and want to create a group of three, we have

n = 12

and

r = 3 .

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

SubstituteII

$n = 12$ , $r = 3$

_{12} C_{3} = \frac{1 2 !}{( 1 2 - 3 ) ! 3 !}

▼

Evaluate right-hand side

SubTerm

Subtract term

_{12} C_{3} = \frac{1 2 !}{9 ! 3 !}

Write as a product

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0 \cdot 9 !}{9 ! \cdot 3 !}

CancelCommonFac

Cancel out common factors

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0 \cdot 9 !}{9 ! \cdot 3 \cdot 2 !}

SimpQuot

Simplify quotient

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0}{3 \cdot 2 !}

$2! = 2$

_{12} C_{3} = \frac{1 2 \cdot 1 1 \cdot 1 0}{3 \cdot 2}

Multiply

_{12} C_{3} = \frac{1 3 2 0}{6}

CalcQuot

Calculate quotient

_{12} C_{3} = 220

There are

220

ways we can choose the three person dart team.

Extra

Difference Between Permutations and Combinations

To illustrate the difference between a permutation and a combination, we will consider a selection of $2$ items from a data set of $9 .$

If we are counting the number of permutations, we can switch the order in which the two items were selected which means we have two permutations.

As for the number of combinations, we only have one since the order in which the items are selected does not matter.

Alternative Solution

Using a Graphing Calculator

To calculate the number of combinations, we can also use the built-in $_{n} C_{r}$ function on our graphing calculator. First we enter the number of people in the group in the calculator window, which is $12 .$

Next, push the $MATH$ button, scroll to PRB, and choose the third option. Having chosen nCr, option, we must finish by entering the number of people in the selected group, which is $3 .$

Note that we obtained exactly the same result as when calculating by using the formula for the number of combinations.

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Alternative Solution