Solution: 2018 Winter Final - 5

Author: Michiel Smid

Question

Let $m \geq 2$ and $n \geq 2$ be integers. What does $$ {m \choose 2} + {n \choose 2} + m \cdot n $$ count?

(a)

None of the above.

(b)

The number of ways to choose an ordered pair of 2 elements from a set consisting of $m + n$ elements.

(c)

The number of ways to choose a 2-element subset from a set consisting of $m + n$ elements.

(d)

The number of ways to choose a subset from a set consisting of $m + n$ elements.

Solution

The answer is C, and I will show how the expression in the question shows the number of ways to choose a 2-element subset from a set consisting of m + n elements:

From a set of $m + n$ elements, we can form 2-element subsets in 3 different ways:

Choose 2 elements from the $m$ set: there are $\binom{m}{2}$ ways to do this
Choose 2 elements from the $n$ set: there are $\binom{n}{2}$ ways to do this
Choose 1 element from the $m$ set and 1 element from the $n$ set: there are $\binom{m}{1} \cdot \binom{n}{1} = m \cdot n$

Now, if we apply the Sum Rule to all of these cases above, we get $\binom{m}{2} + \binom{n}{2} + m \cdot n$, which matches the expression above.