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Question: If \[^n{P_4} = 12.\left( {^n{P_2}} \right)\], find n. (1) 6 (2) 8 (3) 10 (4) -2...

If nP4=12.(nP2)^n{P_4} = 12.\left( {^n{P_2}} \right), find n.
(1) 6
(2) 8
(3) 10
(4) -2

Explanation

Solution

Here we will simply use the formula for permutation and sole for the value of n.
The formula for permutation or arrangement of r numbers out of n numbers is given by:-
nPr=n!(n−r)!^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 out of n terms.

Complete step-by-step answer:
The given equation is: -
nP4=12.(nP2)^n{P_4} = 12.\left( {^n{P_2}} \right)…………………………….. (1)
Now we know that,
The formula for permutation or arrangement of r numbers out of n numbers is given by:-
nPr=n!(n−r)!^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 out of n terms.
Hence, the value of nP4^n{P_4} is given by:-
nP4=n!(n−4)!^n{P_4} = \dfrac{{n!}}{{\left( {n - 4} \right)!}}……………………… (2)
Now, the value of nP2^n{P_2} is given by:-
nP2=n!(n−2)!^n{P_2} = \dfrac{{n!}}{{\left( {n - 2} \right)!}} ………………………. (3)
Putting the values from equation 2 and 3 in equation 1 we get:-
n!(n−4)!=12.(n!(n−2)!)\dfrac{{n!}}{{\left( {n - 4} \right)!}} = 12.\left( {\dfrac{{n!}}{{\left( {n - 2} \right)!}}} \right)
Now we know that,
n!=n×(n−1)!×....................×1n! = n \times \left( {n - 1} \right)! \times .................... \times 1
Hence,
(n−2)!=(n−2)×(n−3)×(n−4)!\left( {n - 2} \right)! = \left( {n - 2} \right) \times \left( {n - 3} \right) \times \left( {n - 4} \right)!
Putting this value in above equation we get:-
n!(n−4)!=12.(n!(n−2)×(n−3)×(n−4)!)\dfrac{{n!}}{{\left( {n - 4} \right)!}} = 12.\left( {\dfrac{{n!}}{{\left( {n - 2} \right) \times \left( {n - 3} \right) \times \left( {n - 4} \right)!}}} \right)
Cancelling the required terms we get:-
1=12.(1(n−2)×(n−3))1 = 12.\left( {\dfrac{1}{{\left( {n - 2} \right) \times \left( {n - 3} \right)}}} \right)
Simplifying it we get:-
(n−2)(n−3)=12\left( {n - 2} \right)\left( {n - 3} \right) = 12
Solving it further we get:-
n2−2n−3n+6=12{n^2} - 2n - 3n + 6 = 12
⇒n2−5n−6=0\Rightarrow {n^2} - 5n - 6 = 0
Now solving the above equation by middle term split we get:-
n2−6n+n−6=0{n^2} - 6n + n - 6 = 0
Now taking the terms we get:-
n(n−6)+1(n−6)=0n\left( {n - 6} \right) + 1\left( {n - 6} \right) = 0
⇒(n−6)(n+1)=0\Rightarrow \left( {n - 6} \right)\left( {n + 1} \right) = 0
Solving for n we get:-
n−6=0;n+1=0n - 6 = 0;n + 1 = 0
⇒n=6;n=−1\Rightarrow n = 6;n = - 1
Now since the value of n cannot be negative therefore, we will ignore the value -1
Hence, the value of n is 6.

Therefore, option (1) is the correct option.

Note: In such questions we directly have to apply the formula and then find the equation and solve that equation to get the desired solution.
Also, students should note that the value of n as well as r cannot be negative in the following formula:-
nPr=n!(n−r)!^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}