The number of solutions $(x, y, z)$ to the equation $x - y - z = 25$, where $x, y$, and $z$ are positive integers such that $x \le 40, y \le 12$, and $z \le 12$ is

We need to find how many sets of positive integers $(x, y, z)$ satisfy:

$x - y - z = 25$

$x \leq 40$, $y \leq 12$, $z \leq 12$

Since we're looking for positive integers, we know that $x \geq 1$, $y \geq 1$, and $z \geq 1$.

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From the equation $x - y - z = 25$:

$x = 25 + y + z$

This is our fundamental relationship that we'll use throughout the solution.

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Since $y \geq 1$ and $z \geq 1$ (positive integers), the minimum value of $x$ is:

$x_{min} = 25 + 1 + 1 = 27$

Since $x \leq 40$ (given constraint), we have:

$27 \leq x \leq 40$

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From $x = 25 + y + z$ and $x \leq 40$:

$25 + y + z \leq 40$

$y + z \leq 15$

Combined with our other constraints:

$1 \leq y \leq 12$

$1 \leq z \leq 12$

$y + z \leq 15$

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For each value of $x$, we need $y + z = x - 25$.

Once we fix $x$, the equation becomes $y + z = \text{constant}$, and we just need to count how many ways we can write this constant as a sum of two positive integers within our bounds.

Let's count systematically:

For $x = 27$: $y + z = 2$, valid pairs $(y,z)$ are $(1,1)$, giving us 1 solution

For $x = 28$: $y + z = 3$, valid pairs $(y,z)$ are $(1,2), (2,1)$, giving us 2 solutions

For $x = 29$: $y + z = 4$, valid pairs $(y,z)$ are $(1,3), (2,2), (3,1)$, giving us 3 solutions

This pattern continues until:

For $x = 38$: $y + z = 13$, valid pairs $(y,z)$ are $(1,12), (2,11), \ldots, (12,1)$, giving us 12 solutions

For $x = 39$: $y + z = 14$, valid pairs $(y,z)$ are $(2,12), (3,11), \ldots, (12,2)$, giving us 11 solutions

For $x = 40$: $y + z = 15$, valid pairs $(y,z)$ are $(3,12), (4,11), \ldots, (12,3)$, giving us 10 solutions

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We need to sum up all the counts:

$\text{Total} = 1 + 2 + 3 + \ldots + 12 + 11 + 10$

First part: $1 + 2 + 3 + \ldots + 12 = \dfrac{12 \times 13}{2} = 78$

Second part: $11 + 10 = 21$

Therefore: $\text{Total} = 78 + 21 = 99$

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The number of solutions is 99.

Our answer makes sense because we're essentially counting lattice points (integer coordinate points) in a bounded region, and our systematic approach ensures we don't miss any valid combinations.