Solution: 2019 Winter Final - 3

Author: Michiel Smid

Question

Consider a set $S$ consisting of 25 beer bottles $b_1,b_2,...,b_{25}$ and 30 cider bottles $c_1,c_2,...,c_{30}$. How many 10-element subsets of $S$ contain at least 2 beer bottles?

(a)

${55 \choose 10} - {30 \choose 10} - 25 \cdot {29 \choose 9}$

(b)

${55 \choose 10} - {30 \choose 10} - {30 \choose 9}$

(c)

${55 \choose 10} - {30 \choose 10} - 25 \cdot {30 \choose 9}$

(d)

${55 \choose 10} - {30 \choose 10} - 25 \cdot {30 \choose 10}$

Solution

Find S

Let S be the set of all possible 10-element subsets of $S$

We choose 10 bottles from the total of 55 bottles: $\binom{55}{10}$
Find A

Let A be the set of all possible 10-element subsets of $S$ that contain 0 beer bottles

We choose 10 bottles from the 30 cider bottles: $\binom{30}{10}$
Find B

Let B be the set of all possible 10-element subsets of $S$ that contain 1 beer bottle

We choose 1 beer bottle from the 25 beer bottles and 9 cider bottles from the 30 cider bottles: $\binom{25}{1} \cdot \binom{30}{9}$
Profit

$ |S| - |A| - |B| $

$= \binom{55}{10} - \binom{30}{10} - \binom{25}{1} \cdot \binom{30}{9} $

$ = \binom{55}{10} - \binom{30}{10} - 25 \cdot \binom{30}{9} $