Count the number of integer solutions to $x_1+x_2+x_3+x_4+x_5+x_6=25$ when, $x_1,x_2,x_3$ are odd and $x_4,x_5,x_6$ are even and $x_i is N$

Question

How to count the number of integer solutions to $$x_1+x_2+x_3+x_4+x_5+x_6=25$$

When, $x_1,x_2,x_3$ are odd ($2k+1$) and $x_4,x_5,x_6$ are even ($2k$), $x_i \in N$.

Is there a general formula to calculate things like this?

Answer 1

Let $x_i=2a_i+1(i=1,2,3)$ and $x_i=2a_i+2(i=4,5,6)$ because all of $x_i$ is integer.

Then the equation changes like this: $a_1+a_2+\cdots+a_6=8$.

The number of ordered pair $(a_1,a_2,\cdots,a_6)$ is $ _6\mathrm H_8={{6+8-1}\choose8}={13\choose8}=1287$.

Answer 2

The sum of three odd numbers is odd, and the sum of three even numbers is even. Therefore, the sum of three odd numbers and three even numbers must be odd and can not be $26$. So, there is no solution with the given conditions.