Find positive integers $a, b$ such that $491! = 11^a \cdot 7^b \cdot c $

Following the hints given in the comments, since $11\cdot 44=484$, for $491!$ the factor $11$ appears at least $44$ times and we need to add

• one more factor for $11\cdot 44$
• one more factor for $11\cdot 33$
• one more factor for $11\cdot 22$
• one more factor for $11\cdot 11$

therefore the exponent for $11$ is $48$.

Multiples of $11$ are $\{11,22,33,...,484\}$ which are $44$ of them and among them we have multiples of $121$ which are $\{121,242,363,484\}$ which are four of them.

Thus the power of $11$ should be $44+4=48$

With $7$ we have multiples of $7$ as $\{7,14,21,...,490\}$ which are $70$ of them.

Among them we have multiples of $49$ which are $\{ 49,98,...,490\}$ which are $10$ of them.

Among these we have one multiple of $343$ which is $\{343\}$

Thus the power of $7$ is $$70+10+1$$ which is $81$

For $a$, we first count the number of multiples of $11$, which are $\left[\dfrac{491}{11}\right]=44$. However, the multiples of $11^2$ give a more "$11$", so we need to count again: $\left[\dfrac{491}{11^2}\right]=4$. Therefore, $a=44+4=48$.

Similarly, $b=\left[\dfrac{491}{7}\right]+\left[\dfrac{491}{7^2}\right]+\left[\dfrac{491}{7^3}\right]=70+10+1=81$.

Actually, if you want to find the power of a prime $p$ of the prime factorization of $n!$, you can use this summation: $$\sum_{k=1}^\infty \left[\dfrac{n}{p^k}\right]$$