Example 42

EXAMPLE 42 Calculating numbers of combinations

Find the following numbers of combinations:

Solution

$_{6} C_{2} = \frac{6!}{2! (6 - 2)!} = \frac{6 \cdot 5 \cdot 4!}{2 \cdot 1 \cdot 4!} = \frac{30}{2} = 15$
$_{6} C_{3} = \frac{6!}{3! (6 - 3)!} = \frac{6 \cdot 5 \cdot 4 \cdot 3!}{3 \cdot 2 \cdot 1 \cdot 3!} = \frac{120}{6} = 20$
$_{6} C_{4} = \frac{6!}{4! (6 - 4)!} = \frac{6!}{(6 - 4)! 4!} = \frac{6 \cdot 5 \cdot 4!}{2 \cdot 1 \cdot 4!} = \frac{30}{2} = 15$

NOW YOU CAN DO

Exercises 33–40.