How do you Simplify $\dfrac{\left( n+3 \right)!}{n!}$?

Hint: Now consider the given expression. We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ...3\times 2\times 1$ hence using this expansion we will first expand the numerator till we get n! in the numerator. Now we will cancel the term n! in numerator and denominator and hence we get the required value of the given expression.

Complete step by step solution:
Now to simplify the given expression we will first understand the concept of factorial.
The term $n!$ is read as n factorial.
Now the term $n!$ can be expanded as $n\times \left( n-1 \right)\times \left( n-2 \right)\times ...\times 3\times 2\times 1$ hence using this we can easily find the value of n! for a number n.
Now also note that we have $n!=n\times \left( n-1 \right)!$
Let us understand this with an example.
Consider the number 5!
Now by using the definition of factorial we expand it as 5 × 4 × 3 × 2 × 1.
Hence we get 5! = 120.
Now similarly we can find the factorial for any number.
Now let us consider the given terms $\dfrac{\left( n+3 \right)!}{n!}$
Now we will expand the numerator in the expression till we get n! in the numerator.
Hence we get, $\dfrac{\left( n+3 \right)\times \left( n+2 \right)\times \left( n+1 \right)\times n!}{n!}$
Now cancelling the term n! from numerator and denominator we get the expression as,
$\Rightarrow \dfrac{\left( n+3 \right)!}{n!}=\left( n+3 \right)\times \left( n+2 \right)\times \left( n+1 \right)$

Hence the value of the given expression is $\left( n+3 \right)\times \left( n+2 \right)\times \left( n+1 \right)$ .

Note: Now factorial is used widely in calculating the number of permutations and the number of arrangements. Now n! gives us the number of ways n objects can be arranged. Now let us say we want to select m objects out of n then the number of ways to do this is given by $\dfrac{n!}{\left( n-m \right)!m!}$ and the number is denoted by $^{n}{{C}_{m}}$ .