question about combinatorical

Question

here is task

How many ways are there to choose from the set {1, 2, . . . , 100} three distinct
numbers so that their sum is even?

first of all sum of three numbers is even if only if

1.all number is even
2.two of them is odd and one is even

i know that

(n)   =  n!/(k!*(n-k)!   
(k)   

and can anybody help me to solve this problem

Answer 1

Wouldn't that simply be

(50 choose 3) + (50 choose 2) * (50 choose 1)

?