Source: 2018 Canadian Open Math Challenge Part B Problem 4
-----

Determine the number of $5$-tuples of integers $(x_1,x_2,x_3,x_4,x_5)$ such that
$\text{(a)}$ $x_i\ge i$ for $1\le i \le 5$;
$\text{(b)}$ $\sum_{i=1}^5 x_i = 25$.
1. We start with the given conditions:
   \[
   x_i \ge i \quad \text{for} \quad 1 \le i \le 5
   \]
   and
   \[
   \sum_{i=1}^5 x_i = 25.
   \]

2. To simplify the problem, we perform a change of variables. Let:
   \[
   y_i = x_i - i \quad \text{for} \quad 1 \le i \le 5.
   \]
   This transformation ensures that \( y_i \ge 0 \) because \( x_i \ge i \).

3. Substituting \( y_i = x_i - i \) into the sum condition, we get:
   \[
   \sum_{i=1}^5 y_i = \sum_{i=1}^5 (x_i - i) = \sum_{i=1}^5 x_i - \sum_{i=1}^5 i.
   \]

4. We know that:
   \[
   \sum_{i=1}^5 x_i = 25 \quad \text{and} \quad \sum_{i=1}^5 i = 1 + 2 + 3 + 4 + 5 = 15.
   \]

5. Therefore:
   \[
   \sum_{i=1}^5 y_i = 25 - 15 = 10.
   \]

6. Now, we need to find the number of nonnegative integer solutions to the equation:
   \[
   y_1 + y_2 + y_3 + y_4 + y_5 = 10.
   \]

7. This is a classic "stars and bars" problem, where we need to distribute 10 indistinguishable stars (units) into 5 distinguishable bins (variables \( y_i \)).

8. The number of ways to do this is given by the binomial coefficient:
   \[
   \binom{10 + 5 - 1}{5 - 1} = \binom{14}{4}.
   \]

9. Calculating the binomial coefficient:
   \[
   \binom{14}{4} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001.
   \]

Conclusion:
\[
\boxed{1001}
\]