10. How many ordered quadruples $(a, b, c, d)$ of positive odd integers are there that satisfy the equation $a+b+c+2 d=15 ?$
Answer: 34
Solution: Using the substitution $(a, b, c, d)=\left(2 a_{1}+1,2 b_{1}+1,2 c_{1}+1,2 d_{1}+1\right)$ where $a_{1}, b_{1}, c_{1}, d_{1}$ are nonnegative integers, the problem is equivalent to finding the number $N$ of nonnegative integer solutions of $2 a_{1}+1+2 b_{1}+1+2 c_{1}+1+4 d_{1}+2=15$ or $a_{1}+b_{1}+c_{1}+2 d_{1}=5$. Note that $d_{1} \leq 2$, and for each fixed $d_{1} \in\{0,1,2\}$, the number of nonnegative integer solutions of $a_{1}+b_{1}+c_{1}=5-2 d_{1}$ is $\binom{7-2 d_{1}}{2}$. Hence
$$
N=\sum_{d_{1}=0}^{2}\binom{7-2 d_{1}}{2}=\binom{7}{2}+\binom{5}{2}+\binom{3}{2}=21+10+3=34
$$