How do you simplify $\dfrac{15!}{9!6!}+\dfrac{7!}{10!5!}$ ?

Hint: We can clearly see that this question is from permutations and combinations. We should know what the factorial of a number is. In mathematics, the factorial of a positive integer $n$ is the product of all positive integers less than or equal to $n$. The factorial of a number is often denoted by $n!$.
We can say that $n!=n\times \left( n-1 \right)\times \left( n-2 \right).......3\times 2\times 1$. So let us use this concept to find the value of the given question.

Complete step by step solution:
We know that factorial of a positive integer $n$ is the product of all positive integers less than or equal to $n$. The factorial of a number is often denoted by $n!$.
In mathematical terms, we can represent in the following way :
$\Rightarrow n!=n\times \left( n-1 \right)\times \left( n-2 \right).......3\times 2\times 1$.
So we have $15!$. So we have to keep multiplying it and reducing it by $1$ until we get $1$.
We have $9!$. So we have to keep multiplying it and reducing it by $1$ until we get $1$.
We have $6!$. So we have to keep multiplying it and reducing it by $1$ until we get $1$.
And then we are supposed to multiply $9!$ with $6!$. And then we should divide $15!$ with what we get.
We have $7!$. So we have to keep multiplying it and reducing it by $1$ until we get $1$.
We have $10!$. So we have to keep multiplying it and reducing it by $1$ until we get $1$.
We have $5!$. So we have to keep multiplying it and reducing it by $1$ until we get $1$.
And then we are supposed to multiply $10!$ with $5!$. And then we should divide $7!$ with what we get.
And then add both of these values.
Let us do them.
$\begin{align}
  & \Rightarrow \dfrac{15!}{9!6!}+\dfrac{7!}{10!5!} \\
 & \Rightarrow \dfrac{15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times 6\times 5\times 4\times 3\times 2\times 1}+\dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times 5\times 4\times 3\times 2\times 1} \\
 & \Rightarrow 7\times 13\times 11\times 5+\dfrac{1}{10\times 9\times 8\times 5\times 4\times 3\times 2\times 1} \\
 & \Rightarrow 5005+\dfrac{1}{86400} \\
 & \Rightarrow 5005+0.000011=5005.000011 \\
\end{align}$
$\therefore $ Upon simplifying $\dfrac{15!}{9!6!}+\dfrac{7!}{10!5!}$, we get $5005.000011$.

Note: We should be careful while solving as there is a lot of scope of calculation errors. We use this concept a lot in permutation and combination. We should know the intuition behind factorial. We should know why we are using it and where it's needed. It is a tricky chapter. A lot of practice is required to crack the logic behind each question and do them quickly and accurately in the exam. We should understand the logic behind each question we solve.