Problem: The number of different 10-letter "words" that can be made from the letters of the word REASSESSES is the same as the number of different $x$-letter "words" that can be made from the letters of the word REDUCTIONS. Compute $x$. Solution: There are $\frac{10!}{5!3!} = 10 \cdot 9 \cdot 8 \cdot 7$ such 10-letter words that can be made from the word REASSESSES (the idea is we order this $10!$ and then, for each ordering, divide out the permutaitons of the five letters $S$ and three letters $E$). What do you know, we got lucky because REDUCTIONS is $10$ letters and the letters are all distinct, so $10 \cdot 9 \cdot 8 \cdot 7$ denotes all $4$-letter words that can be made from REDUCTIONS.