How do you simplify the given division: $\dfrac{101!}{99!}$?

Hint: We start solving the problem by equating the given division to a variable. We then recall the fact that $n!$ is defined as the multiplication of all the natural numbers that were less than or equal to n i.e., $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ....\times 3\times 2\times 1$. We use this result and then cancel the terms that were commonly present in both numerator and denominator to proceed through the problem. We then make the necessary calculations to get the required result.

Complete step by step answer:
According to the problem, we are asked to simplify the given division: $\dfrac{101!}{99!}$.
Let us assume $d=\dfrac{101!}{99!}$ ---(1).
We know that $n!$ is defined as the multiplication of all the natural numbers that were less than or equal to n i.e., $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ....\times 3\times 2\times 1$. Let us use this result in equation (1).
$\Rightarrow d=\dfrac{101\times 100\times 99\times 98\times ......\times 3\times 2\times 1}{99\times 98\times ......\times 3\times 2\times 1}$ ---(2).
From equation (2), we can see that the numerator and denominator have common multiples $99\times 98\times ......\times 3\times 2\times 1$. So, let us cancel those terms to get the required answer.
$\Rightarrow d=101\times 100$.
$\Rightarrow d=10100$.
So, we have found the simplified result of the given division $\dfrac{101!}{99!}$ as 10100.

$\therefore $ The simplified result of the given division $\dfrac{101!}{99!}$ is 10100.

Note: We should keep in mind that we have to multiply only natural numbers while we are solving problems related to factorial $\left( ! \right)$. We can also solve this problem by making use of the fact that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)!$ which gives the similar answer. Whenever we get this type of problem, we first try to recall the required definitions which lead us to the required answer. Similarly, we can expect problems to find the value of the given term $\dfrac{2!}{0!}$.