Question

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

Answer 1

Hint:
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.