Solution: 2015 Winter Final - 1

Author: Michiel Smid

Question

Consider a set $S$ consisting of 20 integers; 5 of these are even and the other 15 integers in $S$ are odd. What is the number of 7-element subsets of $S$ having exactly 3 even integers?

(a)

${5\choose 3} + {15 \choose 4}$

(b)

${5 \choose 4}{15 \choose 3}$

(c)

${5 \choose 3}{15 \choose 4}$

(d)

${20 \choose 7}$

Solution

First, we choose 3 of the 5 even integers: $ \binom{5}{3} $

Then, we choose 4 of the 15 odd integers: $ \binom{15}{4} $

The total number of 7-element subsets of $S$ having exactly 3 even integers is $ \binom{5}{3} \cdot \binom{15}{4} $