How do you simplify $\dfrac{9!}{7!2!}$ ?

Hint: We know that factorial of any positive integer is equal to multiplication of all the numbers from 1 to that number , so we can write $x!=1\times 2\times .....\times \left( n-1 \right)\times n$ , in the above question we do not have to calculate the value of factorial of 9, we can see that in the denominator $7!$ is there so we will cancel out some numbers

Complete step by step answer:
We have simplify $\dfrac{9!}{7!2!}$
$9!$ is product of all numbers from 1 to 9
We can write $9!$ as $1\times 2\times .....\times 8\times 9$ , similarly $7!$ as product of numbers from 1 to 7
We can write $7!$ as $1\times 2\times .....\times 6\times 7$
So we can see $7!$ is the factor of $9!$
If divide numerator and denominator by $7!$ we will get
$\dfrac{9!}{7!2!}=\dfrac{8\times 9}{2!}$
The value of factorial of 2 is equal to 2
So we can write $\dfrac{9!}{7!2!}=\dfrac{72}{2}=36$
The value of $\dfrac{9!}{7!2!}$ is equal to 36.


Note:
The total combination of r objects from n objects represented as ${}^{n}{{C}_{r}}$ which is equal to $\dfrac{n!}{r!\left( n-r \right)!}$ so if we observe we can see that $\dfrac{9!}{7!2!}$ is equal to ${}^{9}{{C}_{7}}$ or ${}^{9}{{C}_{2}}$ , another method to find the value of $\dfrac{9!}{7!2!}$ or ${}^{9}{{C}_{7}}$ or ${}^{9}{{C}_{2}}$ to use the Pascal’s triangle . The value of ${}^{9}{{C}_{7}}$ will be the third term or seventh term of tenth row of Pascal's triangle. We know that the nth row of Pascal’s triangle contains the term ${}^{n-1}{{C}_{0}}$ , ${}^{n-1}{{C}_{1}}$, ${}^{n-1}{{C}_{2}}$ , …. , ${}^{n-1}{{C}_{n-1}}$ . So there are total n+1 terms in the nth row of Pascal’s triangle