[Combinatorics]
Problem: Suppose that 5-letter "words" are formed using only the letters A, R, M, and L. Each letter need not be used in a word, but each word must contain at least two distinct letters. Compute the number of such words that use the letter A more than any other letter.

Solution: Condition on the number $n$ of A's that appear in the word; $n$ is at least two, because of the requirement that $\mathbf{A}$ occur more often than any other letter, and $n$ is at most 4 , because of the requirement that there be at least two distinct letters. In the case $n=4$, there are 3 choices for the other letter, and 5 choices for where to place it, for a total of 15 possibilities. In the case $n=3$, there are two possibilities to consider: either a second letter occurs twice, or there are two distinct letters besides A. If a second letter occurs twice, there are 3 choices



for the other letter, and $\frac{5 !}{3 ! \cdot 2 !}=10$ ways to arrange the three A's and two non-A's, for their locations, for a total of 30 choices. If there are two distinct letters besides $A$, then there are $\left(\begin{array}{l}3 \\ 2\end{array}\right)=3$ ways to pick the two letters, and $\frac{5 !}{3 ! \cdot 1 ! \cdot 1 !}=20$ ways to arrange them, for a total of 60 words. Thus there are a combined total of 90 words when $n=3$. In the case $n=2$, no other letter can occur twice, so all the letters R, M, L, must appear in the word; they can be arranged in $\frac{5 !}{2 ! \cdot 1 ! \cdot 1 ! \cdot 1 !}=60$ ways. The total number of words satisfying the conditions is therefore $15+90+60=\mathbf{1 6 5}$.

Final answer: 165