B4. Determine the number of 5 -tuples of integers $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ such that
1. $x_{i} \geq i$ for $1 \leq i \leq 5$;
2. $\sum_{i=1}^{5} x_{i}=25$
Solution: Let $y_{i}=x_{i}-i$ for $1 \leq i \leq 5$. Then we wish to find the number of 5 -tuples $\left(y_{1}, y_{2}, y_{3}, y_{4}, y_{5}\right)$ of nonnegative integers such that:
$$
\sum_{i=1}^{5} y_{i}=25-(1+2+3+4+5)=10 \text {. }
$$

Consider $10+5-1=14$ boxes in a row. Cross out $5-1=4$ of them. Starting from the left, let there be $y_{1}$ boxes before the first crossed out one, $y_{2}$ between the first and second crossed out box, continuing to $y_{5}$ boxes between the $4^{\text {th }}$ crossed out box and the end of the row. Then the $y_{i}$ are nonnegative integers, and $\sum_{i=1}^{5} y_{i}=10$. So this is an 5 -tuple that we want! Note that given an 5 -tuple $\left(y_{1}, y_{2}, \ldots, y_{5}\right)$, we cross out boxes according to the above procedure to get a bijection between the two. There are $\binom{14}{4}$ ways of crossing out boxes, hence the number of 5-tuples is $\binom{14}{4}=\frac{14 \times 13 \times 12 \times 11}{2 \times 3 \times 4}=7 \times 13 \times 11=1001$.
The answer is 1001 .