If $(n + 1)! = 12(n - 1)!$, find the value of $n$.

Hint: In this question, we are given an equation in factorials and we have been asked to find the value of $n$. Start by expanding the LHS of the equation until one of the values becomes equal to the given value in RHS. Once you get the same value on both the sides, cancel them. Then shift the RHS to LHS. You will get a quadratic equation. Solve the equation by splitting the middle term and find the value of $n$.

Formula used: $x! = x \times (x - 1) \times (x - 2) \times (x - 3) \times ...... \times 3 \times 2 \times 1$

Complete step-by-step answer:
We are given an equation in factorials. $ \Rightarrow (n + 1)! = 12(n - 1)!$
We can see that If we expand the LHS of the given equation, it will give us a value equal to RHS. Other than this, we have no other way to start this question. So, let us start expanding the LHS.
$ \Rightarrow (n + 1)! = 12(n - 1)!$
$ \Rightarrow (n + 1)(n + 1 - 1)(n + 1 - 2)! = 12(n - 1)!........n! = n(n - 1)!$
Simplifying,
$ \Rightarrow (n + 1)(n)(n - 1)! = 12(n - 1)!$
Now we can see the term $(n - 1)!$ on both the sides. Let us cancel them to get rid of factorials.
$ \Rightarrow (n + 1)(n)\dfrac{{(n - 1)!}}{{(n - 1)!}} = 12$
We will get,
$ \Rightarrow (n + 1)(n) = 12$
Opening the brackets on LHS,
$ \Rightarrow {n^2} + n = 12$
Shifting 12 to the other side to get a quadratic equation.
$ \Rightarrow {n^2} + n - 12 = 0$
Using splitting the middle term to find the value of $n$.
$ \Rightarrow {n^2} + 4n - 3n - 12 = 0$
Simplifying,
$ \Rightarrow n(n + 4) - 3(n + 4) = 0$
$ \Rightarrow (n - 3)(n + 4) = 0$
Finding the value of $n$.
$ \Rightarrow n = 3, - 4$

Since, $n$$ = - 4$ is not defined, $n = 3$ is the correct value of $n$.

Note: 1) The factorial operation is encountered in many areas of mathematics, notably in combinatory, algebra, and mathematical analysis. It’s most basic use counts the possible distinct sequences – the permutations – of $n$ distinct objects: there are $n!$. The factorial function can also be extended to non-integer arguments while retaining its most important properties by defining $x! = \Gamma \left( {x + 1} \right)$. Where $\Gamma $ is the gamma function. This is undefined when $x$ is a negative integer.
2) We discarded the value $n$$ = - 4$ because $( - 4 + 1)! = ( - 3)!$ is not defined.
3) It must be kept in mind that while solving an equation, the number of solutions is equal to the degree of the equation. For example, if the degree of the equation is three, the equation will have three solutions.