[Number Theory]
Problem: Let $T=11$. Let $p$ be an odd prime and let $x, y$, and $z$ be positive integers less than $p$. When the trinomial $(p x+y+z)^{T-1}$ is expanded and simplified, there are $N$ terms, of which $M$ are always multiples of $p$. Compute $M$.

Solution: A general term in the expansion of $(p x+y+z)^{T-1}$ has the form $K(p x)^{a} y^{b} z^{c}$, where $a, b$, and $c$ are nonnegative integers such that $a+b+c=T-1$. Using the "stars and bars" approach, the number of nonnegative integral solutions to $a+b+c=T-1$ is the number of arrangements of $T-1$ stars and 2 bars in a row (the bars act has separators and the " 2 " arises because it is one less than the number of variables in the equation). Thus there are $\left(\begin{array}{c}T+1 \\ 2\end{array}\right)$ solutions. Each term will be a multiple of $p$ unless $a=0$. In this case, the number of terms that are not multiples of $p$ is the number of nonnegative integral solutions to the equation $b+c=T-1$, which is $T$ ( $b$ can range from 0 to $T-1$ inclusive, and then $c$ is fixed). Hence $M=\left(\begin{array}{c}T+1 \\ 2\end{array}\right)-T=\frac{T^{2}-T}{2}$. With $T=11$, the answer is $\mathbf{5 5}$

Final answer: 55