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 ≤ 40, y ≤ 12, and z ≤ 12 is

The correct option is B.

We are given the equation:
x − y − z = 25 ⇒ x − y − z − 25 = 0 ⇒ x = y + z + 25

We are given that x, y, z are positive integers, which means:
x ≥ 1, y ≥ 1, z ≥ 1
Also, we have constraints:
x ≤ 40, y ≤ 12, z ≤ 12

Since x = y + z + 25, the constraint x ≤ 40 becomes:
y + z + 25 ≤ 40 ⇒ y + z ≤ 15

We need to find the number of positive integer solutions (y, z) such that:
y + z ≤ 15, subject to:
1 ≤ y ≤ 12 and 1 ≤ z ≤ 12

Let's define a new variable s = y + z. Since y ≥ 1 and z ≥ 1, the minimum value of s is 2. The maximum value of s is 15. Thus, 2 ≤ s ≤ 15.

For any given sum s = y + z, the number of integer pairs (y, z) where y ≥ 1 and z ≥ 1 is s − 1. However, we must exclude cases where y > 12 or z > 12.

Since s ≤ 15, we cannot have both y > 12 and z > 12 simultaneously (since that would require s ≥ 13 + 13 = 26). Thus, at most one of the variables can exceed 12.

Case 1: Total positive integer solutions to y + z ≤ 15 without the upper bounds y ≤ 12, z ≤ 12.
Using a dummy variable w ≥ 0:
y + z + w = 15, where y ≥ 1, z ≥ 1, w ≥ 0.
Let y' = y − 1 ≥ 0 and z' = z − 1 ≥ 0.
y' + z' + w = 13.
The number of non-negative integer solutions is:
$\left(\begin{array}{}13+3-1\end{array},\begin{array}{}3-1\end{array}\right)=\left(\begin{array}{}15\end{array},\begin{array}{}2\end{array}\right)=\frac{15\times 14}{2}=105$

Case 2: Solutions where y > 12 (i.e., y ≥ 13) and z ≥ 1.
Let y'' = y − 13 ≥ 0 and z' = z − 1 ≥ 0.
y + z ≤ 15 ⇒ (y'' + 13) + (z' + 1) ≤ 15 ⇒ y'' + z' ≤ 1.
With dummy variable w ≥ 0:
y'' + z' + w = 1.
The number of non-negative integer solutions is:
$\left(\begin{array}{}1+3-1\end{array},\begin{array}{}3-1\end{array}\right)=\left(\begin{array}{}3\end{array},\begin{array}{}2\end{array}\right)=3$

Case 3: Solutions where z > 12 (i.e., z ≥ 13) and y ≥ 1.
By symmetry, the number of solutions is also 3.

Subtracting the invalid solutions from the total:
Number of valid solutions = 105 − 3 − 3 = 99.

Therefore, the correct option is B (99).