Question

If nP4 = 12 x nP2, find value of n​

Answer 1

Step-by-step explanation:

npr=n!/(n-r)!

np4 =n!/(n-4)!

...... .............

nP2=n!/(n-2)!

12 =(n-2)!/(n-4)!..... by given data 12 written here

12 =(n-2)(n-3)

n^2-5n+6-12=0

n^2-5n-6=0

n^2-6n+n-6=0

(n-6)(n+1)=0

n=6. or. -1

Answer 2

The value of n is 6.

Step-by-step explanation:

We are given the following in the question:

$^nP_4 =12\times ^nP_2$

Formula:

$^nP_r = \dfrac{n!}{(n-r)!}$

Putting values, we get,

$\dfrac{n!}{(n-4)!}=12\times \dfrac{n!}{(n-2)!}\\\\\dfrac{(n-2)!}{(n-4)!} = 12\\\\\dfrac{(n-2)(n-3)(n-4)!}{(n-4)!} = 12\\\\(n-2)(n-3) = 12\\\Rightarrow n^2-5n-6=0\\\Rightarrow (n-6)(n+1) = 0\\\Rightarrow n = 6, n =-1$

Since n cannot be negative, thus, the value of n is 6.

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