Courses for Kids
Free study material
Offline Centres
Store Icon

If \[{}^n{P_2} = 90\] then find the value of \[n\].

Last updated date: 20th Jun 2024
Total views: 386.1k
Views today: 6.86k
386.1k+ views
Hint: 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
 \[\begin{array}{l} \Rightarrow \left( {n + 9} \right) = 0\\ \Rightarrow n = - 9\end{array}\]
 \[\begin{array}{l} \Rightarrow \left( {n - 10} \right) = 0\\ \Rightarrow n = 10\end{array}\]
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.