# Exercise C.1.4

In how many ways can we choose three distinct numbers from the set $\{1,2,\ldots,99\}$ so that their sum is even?

There are 49 even numbers and 50 odd in that set. To get an even number, we either need two odd and one even or three even.

$$\langle \text{2 odd, 1 even} \rangle = 49\frac{50!}{2! \cdot 48!} = \frac{50 \cdot 49^2}{2} \\ \langle \text{3 even} \rangle = \frac{49!}{3! \cdot 46!} = \frac{49 \cdot 48 \cdot 47}{6} \\ \langle \text{even sum} \rangle = \frac{49 \cdot 48 \cdot 47}{6} + \frac{50 \cdot 49^2}{2} = 78449$$