Number of answers of equation amongs odd natural numbers

Question

How many answer The following Equation has, in set of odd natural numbers?

$x_1+x_2+...+x_k=n$, $k \equiv^2 n$

Solution: Comb ( [(n+k)/2]-1, k-1), comb means combination. how we get this?

Answer 1

If $k$ is odd and $n$ is even, there are no solutions in odd integers. Similarly, if $k$ is even and $n$ is odd, there are no solutions in odd integers. So we may assume that $n$ and $k$ have the same parity (both odd or both even).

The number of solutions of $x_1+\cdots+x_k=n$ in odd positive integers is the number of solutions of $y_1+\cdots+y_k=n+k$ in even positive integers. For we go from odd integers $x_i$ to even positive integers $y_i$ by adding $1$ to each $x_i$, and the process is reversible.

Similar reasoning shows that the number of solutions of $y_1+\cdots+y_k$ in even positive integers is the same as the number of solutions of $z_1+\cdots+z_k=\frac{n+k}{2}$ in positive integers.

By standard Stars and Bars, there are $\binom{\frac{n+k}{2}-1}{k-1}$ solutions of $z_1+\cdots+z_k=\frac{n+k}{2}$ in positive integers.

Remark: Alternately, the number of solutions of $x_1+\cdots+x_k=n$ is the number of solutions of $v_1+\cdots+v_k=n-k$ in non-negative even integers (subtract $1$ from each $x_i$, which is the same as the number of solutions of $w_1+\cdots+w_k=\frac{n-k}{2}$ in non negative integers. Stars and Bars deals with this problem.