A seemingly easy problem

[Darya]

In how many ways can we select 3 different numbers from 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 so that their sum is an even number?

I thought it could be even+even+even and even+odd+odd. So the number of ways is 5*4*3+5*5*4=160. The correct answer is 60. What am I doing wrong? Thanks!

 

[Dr.Peterson]

You are counting ways in which order counts; but this is not a permutation question. Think in terms of combinations (one case at a time). How many ways are there to choose three of the even numbers? How many ways to choose one of the even numbers and two of the odd numbers?

 

[Darya]

You are counting ways in which order counts; but this is not a permutation question. Think in terms of combinations (one case at a time). How many ways are there to choose three of the even numbers? How many ways to choose one of the even numbers and two of the odd numbers?

Ohhhh, right! Then it's 50+10=60. Thank you!