Question

np2=20 find value of n permutations
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Answer 1

Step-by-step explanation:

nP2=20

n!/(n-2)! = 20

n(n-1)(n-2)!/(n-2)! = 20

n(n-1)=20

n=20 or n-1=20 =>> n=21

Answer 2

The value of n is 5.

Step-by-step explanation:

Given : $^nP_2=20$

To find : The value of n permutations ?

Solution :

We know the formula of permutation is $^nP_r=\frac{n!}{(n-r)!}$

Write the expression $^nP_2=20$ as,

$\frac{n!}{(n-2)!}=20$

$\frac{n(n-1)(n-2)!}{(n-2)!}=20$

$n(n-1)=20$

$n^2-n=20$

$n^2-n-20=0$

$n^2-5n+4n-20=0$

$n(n-5)+4(n-5)=0$

$(n-5)(n+4)=0$

$n=5,-4$

Reject n=-4.

So, the value of n is 5.

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