Question

If ${}^n{P_4} = 12\left[ {{}^n{P_2}} \right]$, find $n$.
A) 6
B) 8
C) 10
D) $- 2$

Answer 1

Here we need to find the value of $n$. We will use the formula of permutations so as to expand both the sides of the equation. Permutation is a way of arranging elements of a set in a certain order and sequence. We will evaluate the factorial and cancel out the similar terms. Solving further will give a quadratic equation which we will solve using factorization to find the value of $n$.
Formula Used: We will use the formula of permutation ${}^n{P_r} = \dfrac{{n!}}{{\left[ {n - r} \right]!}}$ to expand, where $r$ elements are arranged from $n$ number of sets.

Complete step by step solution:
Here, we will expand both the sides of the equation using the formula ${}^n{P_r} = \dfrac{{n!}}{{\left[ {n - r} \right]!}}$.
We will substitute $r = 4$ for the right hand side of the equation ${}^n{P_4} = 12\left[ {{}^n{P_2}} \right]$ and for the left hand side, we will substitute $r = 2$. After substituting, we will get
$\dfrac{{n!}}{{\left[ {n - 4} \right]!}} = 12\left[ {\dfrac{{n!}}{{\left[ {n - 2} \right]!}}} \right]$
A factorial of a number is the product of all the numbers equal to or less that that number. We can rewrite the above obtained factorial as,
$\dfrac{{n!}}{{\left[ {n - 4} \right]!}} = 12\left[ {\dfrac{{n!}}{{\left[ {n - 2} \right]\left[ {n - 3} \right]\left[ {n - 4} \right]!}}} \right]$
We will now cancel out the common factors from both the sides of the equation.
$1 = 12\left[ {\dfrac{1}{{\left[ {n - 2} \right]\left[ {n - 3} \right]}}} \right]$
Multiplying both the sides of the equation with $\left[ {n - 2} \right]\left[ {n - 3} \right]$, we get
simplify further so as to obtain a quadratic equation.
$\left[ {n - 2} \right]\left[ {n - 3} \right] = 12$
Multiplying the terms, we get
$\Rightarrow {n^2} - 2n - 3n + 6 = 12$
Adding and subtracting the like terms, we get
$\begin{array}{l} \Rightarrow {n^2} - 5n + 6 - 12 = 0\\\ \Rightarrow {n^2} - 5n - 6 = 0\end{array}$
Here, we will use factorization to solve the quadratic equation. We will express $- 5$ as the sum of the factors of the number $- 10$. This will be done in the following manner,
$\begin{array}{l}{n^2} - 5n - 6 = 0\\\ \Rightarrow {n^2} - 6n + n - 6 = 0\end{array}$
Now, common out factors from two terms when taken together. This will leave us with two terms as a product of each other.
$\begin{array}{l}{n^2} - 6n + n - 6 = 0\\\ \Rightarrow n\left[ {n - 6} \right] + 1\left[ {n - 6} \right] = 0\\\ \Rightarrow \left[ {n + 1} \right]\left[ {n - 6} \right] = 0\end{array}$
Now, as we can see the product of the two terms is equal to 0. But logically, this is only possible when either of the terms is 0 as only the product of a number and 0 will give 0.
This implies that either $n + 1$ or $n - 6$ is 0. Equating both of them to 0, we will get,
$\begin{array}{l}n + 1 = 0\\\ \Rightarrow n = - 1\end{array}$
or
$\begin{array}{l}n - 6\\\ \Rightarrow n = 6\end{array}$
As we know that $n$ can only be a positive integer because factorials can only be expressed for positive integers. So, we reject $n = - 1$ as it is negative, and hence, the value of $n$ will be 6.

$\therefore$ Option (1) is the correct answer.

Note:
We can also derive the value of $n$ by using the formula $n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a$ is the coefficient of ${n^2}$, $b$ is the coefficient of $n$ and $c$ is the constant.
For the equation ${n^2} - 5n - 6 = 0$ , substitute $a = 1$, $b = - 5$ and $c = - 6$ in the formula $n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to determine $n$.
After substituting, we will get,
$n = \dfrac{{ - \left[ { - 5} \right] \pm \sqrt {{{\left[ { - 5} \right]}^2} - 4\left[ 1 \right]\left[ { - 6} \right]} }}{{2\left[ 1 \right]}}$
Simplifying the above equation, we get
$\begin{array}{c}n = \dfrac{{5 \pm \sqrt {25 + 24} }}{2}\\\ = \dfrac{{5 \pm \sqrt {49} }}{2}\end{array}$
Simplifying further, we get two values of $n$.
$n = \dfrac{{5 \pm 7}}{2}\\\ = \dfrac{{5 + 7}}{2},\dfrac{{5 - 7}}{2}\\\ = \dfrac{{12}}{2},\dfrac{{ - 2}}{2}\\\ = 6, - 1$
As $n$ can never be negative so the value of $n$ will be 6.