# Thread: Another easy combinatorics problem

1. ## Another easy combinatorics problem

How many three-digit numbers are there in which the sum of the digits is even?

There are 900 three-digit numbers.
I guess half of them (450) will have even sum of digits. Is this right? I don't know how to prove this, though.

2. ## Re: Another easy combinatorics problem

Originally Posted by alexmahone
How many three-digit numbers are there in which the sum of the digits is even?
There are 900 three-digit numbers.
If three integers add to an even then all three are even or one even and two are odd. Count them.

3. ## Re: Another easy combinatorics problem

Originally Posted by Plato
If three integers add to an even then all three are even or one even and two are odd. Count them.
All three digits are even: 4*5*5 = 100
First digit is even, second and third are odd: 4*5*5 = 100
First digit is odd, second is even, third is odd: 5*5*5 = 125
First digit is odd, second is odd, third is even: 5*5*5 = 125

Total: 100 + 100 + 125 + 125 = 450

4. ## Re: Another easy combinatorics problem

Originally Posted by alexmahone
All three digits are even: 4*5*5 = 100
First digit is even, second and third are odd: 4*5*5 = 100
First digit is odd, second is even, third is odd: 5*5*5 = 125
First digit is odd, second is odd, third is even: 5*5*5 = 125

Total: 100 + 100 + 125 + 125 = 450
Correct, half of them give an even sum.