Question

If ${}^n{P_2} = 90$ then find the value of $n$.

Answer 1

Here, we will compare the given equation to the formula of Permutations and solving this further using factorials we will get a quadratic equation. We will solve the equation using the method of middle term splitting and find the required value of $n$.

Formula Used:
We will use the following formulas:
1. ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, where $n$ is the total number of terms and $r$ is the number of terms to be arranged among them.
2. $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$

Complete step-by-step answer:
We will consider the left hand side of the given equation ${}^n{P_2} = 90$.
Substituting $r = 2$ in the formula ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, we get,
${}^n{P_2} = \dfrac{{n!}}{{\left( {n - 2} \right)!}}$…………………………. $\left( 1 \right)$
Substituting ${}^n{P_2} = 90$ in the above equation, we get
$\Rightarrow \dfrac{{n!}}{{\left( {n - 2} \right)!}} = 90$
Computing the factorials using the formula $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$, we get
$\Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!}} = 90$
Solving the above equation further, we get
$\Rightarrow n\left( {n - 1} \right) = 90$
Multiplying the terms using distributive property, we get
$\Rightarrow {n^2} - n - 90 = 0$
The above equation is a quadratic, we will factorize the equation to find the value of $n$.
Now, splitting the middle term split, we get
$\Rightarrow {n^2} - 10n + 9n - 90 = 0$
Now factoring out common terms, we get
$\Rightarrow n\left( {n - 10} \right) + 9\left( {n - 10} \right) = 0$
Again factoring out the common terms, we get
$\Rightarrow \left( {n + 9} \right)\left( {n - 10} \right) = 0$
Now using zero product property, we get
$\begin{array}{l} \Rightarrow \left( {n + 9} \right) = 0\\\ \Rightarrow n = - 9\end{array}$
Or
$\begin{array}{l} \Rightarrow \left( {n - 10} \right) = 0\\\ \Rightarrow n = 10\end{array}$
But, the total number of terms cannot be negative.
Hence, rejecting the negative value, $n \ne - 9$
Therefore, the required value of $n$ is 10.

Note: In this question, we are required to use the formula of Permutations. Permutation is a way or method of arranging elements from a given set of elements, where the order or sequence of arrangement matters. We might make a mistake by using the formula of combination instead of permutation. Combination is a method of selecting elements from a given set but the order of selection doesn’t matter.