Suppose we have $$x_1 < ... < x_{10}$$ where each $x_i \in \Bbb N$. Specifically, suppose $$x_1,...,x_5 \in 2\Bbb N$$ and $$x_6,...,x_{10} \in 2\Bbb N+1$$
Let $$\sum_{i=1}^{10} x_i = 271$$.
How many possible combinations of $(x_1,...,x_{10})$ are there?
You can consider counting solutions of $$x_1+x_2+\cdots+x_{10}=271$$ as counting arrangements of $271$ identical dots in a row, with $9$ lines to separate them into the $10$ values $x_1,x_2,\ldots,x_{10}$. To count these arrangements, consider that there are $280$ available positions, and you have to choose $9$ positions to take the lines. The answer to this problem is $C(280,9)$.
BUT in your case you have the additional condition that $x_1,\ldots,x_5$ are even and $x_6,\ldots,x_{10}$ are odd. So, set aside $5$ dots - you now have $266$ dots. Consider these as pairs of dots - you have $133$ pairs. Now arrange these $133$ pairs and $9$ lines in a row as above - there are $C(142,9)$ ways of doing so. Now take your $5$ reserved dots and put one into each of the last five spots - this ensures that $x_1,\ldots,x_5$ are even and $x_6,\ldots,x_{10}$ are odd. Hence:
answer: $C(142,9)$.
