Solution: 2015 Winter Final - 2

Author: Michiel Smid

Question

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

(a)

None of the above.

(b)

${5 \choose 5}{15 \choose 2} + {5 \choose 2}{15 \choose 5} - {5 \choose 5}{15 \choose 5}$

(c)

${5 \choose 5}{15 \choose 2} + {5 \choose 2}{15 \choose 5} + {5 \choose 1}{15 \choose 6} + {5 \choose 0}{15 \choose 7}$

(d)

${20 \choose 7} - {5 \choose 4}{15 \choose 4}$

Solution

Let $A$ be the event that a 7-element subset of $S$ has 5 even integers and 2 odd integers.
First, we choose 5 of the 5 even integers: $ \binom{5}{5} $
Then, we choose 2 of the 15 odd integers: $ \binom{15}{2} $
$ |A| = \binom{5}{5} \cdot \binom{15}{2} $
Let $B$ be the event that a 7-element subset of $S$ has 5 odd integers and 2 even integers.
First, we choose 5 of the 15 odd integers: $ \binom{15}{5} $
Then, we choose 2 of the 5 even integers: $ \binom{5}{2} $
$ |B| = \binom{15}{5} \cdot \binom{5}{2} $
Let $C$ be the event that a 7-element subset of $S$ has 6 odd integers and 1 even integer.
First, we choose 6 of the 15 odd integers: $ \binom{15}{6} $
Then, we choose 1 of the 5 even integers: $ \binom{5}{1} $
$ |C| = \binom{15}{6} \cdot \binom{5}{1} $
Let $D$ be the event that a 7-element subset of $S$ has 7 odd integers.
First, we choose 7 of the 15 odd integers: $ \binom{15}{7} $
$|D| = \binom{15}{7} $

The total number of 7-element subsets of $S$ having at least 5 even integers or at least 5 odd integers is...

$ |A| + |B| + |C| + |D| = \binom{5}{5} \binom{15}{2} + \binom{15}{5} \binom{5}{2} + \binom{15}{6} \binom{5}{1} + \binom{15}{7} $