How many sets of numbers can we find?

Question

We have the following set of numbers $1,2\dots14$.

a) How many subsets with 4 numbers can we find that contain only one even number

b) How many subsets with 4 numbers can we find that have a sum of an even number

As for the a, what I am thinking is , every subset with 3 numbers out of $1,3,5,7,9,11,13 $ is $\binom73=\frac{7!}{3!\cdot4!}=35$

So for every subset that I already have I need to add one number out of $ 2,4,6,8,10,12,14$ which means I have 7 choices for each subset. Therefore $7\cdot35$

My problem is with b

Answer 1

We consider the three cases in which four numbers from the set can sum to an even number.

• All numbers are even: $\binom74=35$ ways
• All numbers are odd: 35 ways (there are seven even and seven odd numbers; the situation is symmetric between the four-odd and four-even cases)
• Two numbers are even and two are odd: $\binom72^2=441$ ways

Adding up, we find $35+35+441=511$ admissible subsets.

Answer 2

The sum of numbers is even when we have even numbers of odds!

Here we can have $4$ odds (and no evens), $2$ odds (and $2$ evens), and $0$ odds (and $4$ evens) to let the sum of numbers be even. $\underline{\text{And}}$ means multiplication.

So we have:

$$\displaystyle{7 \choose 4}{7 \choose 0}+{7 \choose 2}{7 \choose 2}+{7 \choose 0}{7 \choose 4}$$