4. In the equality, digits were replaced with letters: identical digits with identical letters, and different ones with different letters. The result was

$$
Y!\cdot P!\cdot A!=K P O C C
$$

What could the original equality have looked like? Find all the options and prove that there are no others.

$N!$ denotes the product of numbers from 1 to $N$, and by definition, $0!=1$.
Solution. If C is not 0, then the left side does not exceed $24 \cdot 6 \cdot 2<10000$.

Then $\mathrm{C}=0$, which means there are two digits on the left that are not less than 5. That is, the left side is not less than $1! \cdot 5! \cdot 6!$, and it cannot be greater due to the five-digit nature. This results in the only (up to the permutation of Y and A and the replacement $0!=1$ !) solution.