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
$\endgroup$ 12 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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:
$$\displaystyle{7 \choose 4}{7 \choose 0}+{7 \choose 2}{7 \choose 2}+{7 \choose 0}{7 \choose 4}$$
$\endgroup$ 2
