# Problem 3. (3 points)

Misha has 10 cards, each with one letter written on it. He can form $7!=$ 5040 different ten-letter words from them. How many different letters can he have on the cards? (Provide all possible variants and prove that there are no others).
# Answer: $5 ; 4$

Solution: The number of words that Misha can form is equal to $\frac{10!}{a!b!\ldots}$, where the numbers $a, b, \ldots$ are the number of times each letter repeats.

$7!=\frac{10!}{720}$, that is, our task is to represent the number 720 as a factorial or a product of several factorials, each greater than 1. $720=6!$, this is one of the representations. In addition, 720 is divisible by 5, so the other options must include $5!=120$. Thus, $720=5!\cdot 6=5!\cdot 3$. It is easy to see that there are no other ways to represent the number 6 as a product of factorials.

Therefore, we get two options. In one of them, corresponding to the formula $\frac{10!}{6!}$, one letter repeats 6 times, and the other 4 are unique, totaling 5 letters. In the other option, defined by the formula $\frac{10!}{5!3!}$, one letter appears 5 times, another 3 times, and there are 2 more unique letters, totaling 4 different letters.