The correct answer is option (E):
351
Let's consider the expansion of $(x + y + z)^{25}$. We are looking for the number of distinct terms. Each term will be of the form $x^a y^b z^c$ where $a, b, c$ are non-negative integers such that $a + b + c = 25$.
The problem of finding the number of distinct terms is equivalent to finding the number of non-negative integer solutions to the equation $a + b + c = 25$. This is a classic combinatorics problem that can be solved using stars and bars.
Imagine we have 25 "stars" (representing the power) and we want to divide them into 3 groups (representing the variables x, y, and z). We can do this by placing 2 "bars" among the stars. For example:
* \* \* \* \* \* | \* \* \* \* \* \* \* \* \* \* \* \* \* \* | \* \* \* \* \* \* \* \* \* \* corresponds to $x^5 y^{15} z^5$.
* | \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* corresponds to $x^0 y^0 z^{25}$.
* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* \* | | \* corresponds to $x^{25} y^0 z^0$.
So we have a total of 25 stars and 2 bars. The number of ways to arrange these is the same as choosing the positions of the 2 bars among the total $25 + 2 = 27$ positions. This can be calculated using combinations, also denoted as "choose".
The number of solutions is therefore given by $\binom{25 + 3 - 1}{3 - 1} = \binom{25 + 2}{2} = \binom{27}{2}$.
$\binom{27}{2} = \frac{27!}{2!(27-2)!} = \frac{27!}{2!25!} = \frac{27 \times 26}{2 \times 1} = \frac{27 \times 26}{2} = 27 \times 13 = 351$.
Therefore, the number of distinct terms in the expansion of $(x + y + z)^{25}$ is 351.