Solution: 2014 Winter Midterm - 1

Author: Michiel Smid

Question

On a table, you see three types of fruit: apples, bananas, and oranges. There are $m \geq 2$ apples, $n \geq 2$ bananas, and $k \geq 2$ oranges. How many ways are there to choose 7 pieces of fruit, if you must take at least two pieces of each type?

(a)

${m + n + k \choose 7} - {m \choose 2} - {n \choose 2} - {k \choose 2}$

(b)

${m + n + k \choose 7} - (m + n + k)$

(c)

${m \choose 2}{n \choose 2}{k \choose 2}(m + n + k)$

(d)

${m \choose 3}{n \choose 2}{k \choose 2} + {m \choose 2}{n \choose 3}{k \choose 2} + {m \choose 2}{n \choose 2}{k \choose 3}$

Solution

We can break this down into 3 cases.

Case 1: 3 apples, 2 bananas, 2 oranges
$\binom{m}{3} \cdot \binom{n}{2} \cdot \binom{k}{2}$
Case 2: 2 apples, 3 bananas, 2 oranges
$\binom{m}{2} \cdot \binom{n}{3} \cdot \binom{k}{2}$
Case 3: 2 apples, 2 bananas, 3 oranges
$\binom{m}{2} \cdot \binom{n}{2} \cdot \binom{k}{3}$

Thus, the total number of ways is $\binom{m}{3} \cdot \binom{n}{2} \cdot \binom{k}{2} + \binom{m}{2} \cdot \binom{n}{3} \cdot \binom{k}{2} + \binom{m}{2} \cdot \binom{n}{2} \cdot \binom{k}{3}$