Simple probability

we're choosing 4 numbers out 10. So the sample space would be $\operatorname{nCr}\left(10,4\right)$ . There are 3 cases where the sum of 4 chosen numbers is even, all the chosen numbers are even, all are odd or 2 odds and 2 evens. For 2 odds and 2 evens, the amount of ways this could happen is $\operatorname{nCr}\left(5,2\right)^2$ . This come from the fact that there are 5 odd numbers and 5 even numbers and we're choosing two of each. The second and third case is all the chosen numbers are even or odd. So this comes out to $2\operatorname{nCr}\left(5,4\right)$ . We can either choose 4 numbers from the 5 even numbers or 4 numbers from the 5 odd numbers. Or you could think of this as 2 ways to choose either odd or even * $\operatorname{nCr}\left(5,4\right)$ . Sum the cases up we will get $\frac{\operatorname{nCr}\left(5,2\right)^2+2\operatorname{nCr}\left(5,4\right)}{\operatorname{nCr}\left(10,4\right)}$ which equals $\frac{11}{21}$ , 11 + 21 = 32