Solution: 2015 Fall Final - 1

Author: Michiel Smid

Question

Consider a set $S$ consisting of 20 integers; 5 of these are strictly positive and the other 15 integers in $S$ are strictly negative. What is the number of 3-element subsets of $S$ having the property that the product of the 3 elements in the subset is negative?

(a)

${15 \choose 3} + 15 \cdot {5 \choose 2}$

(b)

${15 \choose 3} + {15 \choose 2} \cdot 5 + 15 \cdot {5 \choose 2}$

(c)

${20 \choose 3}$

(d)

${15 \choose 3}$

Solution

There are 2 cases to consider:

Let A be the event that the subset contains 3 negative integers.
We choose 3 of the 15 negative integers: $ \binom{15}{3} $
$ |A| = \binom{15}{3} $
Let B be the event that the subset contains 1 negative integer and 2 positive integers.
We choose 1 of the 15 negative integers: $ \binom{15}{1} = 15 $
We choose 2 of the 5 positive integers: $ \binom{5}{2} $
$ |B| = 15 \times \binom{5}{2} $

The total number of 3-element subsets of $S$ having the property that the product of the 3 elements in the subset is negative is the \sum of the subsets in sets $ A \cup B $:

$ |A| + |B| $

$ = \binom{15}{3} + 15 \times \binom{5}{2} $