Question

Find the value of n such that nP5 = 42 nP3 , n>4

Answer 1

Hey mate here is your answer :

• Given that :
nP5 = 42 nP3
n(n-1) (n-2) (n-3) (n-4) = 42 n(n-1) (n-2)

Since n>4 so n(n-1) (n-2) is not equal to 0

(n-3) (n-4) = 43
n²-7n-30 = 0
n²-10n+3n-30 = 0
(n-10) (n+3) = 0
n = 10 or -3

But n cannot be nagative so the value of n is 10

Answer 2

Heya ____

Solution is given here ____

Given info :-

nP5 = 42 nP3

So.....

n(n-1) (n-2) (n-3) (n-4) = 42n(n-1)(n-2)

Since __ n> 4 , so n(n-1) (n-2) is not equals to 0...

Now ____

(n-3)(n-4) = 43

n^2 -7n -30 = 0

n^2 - 10n + 3n -30 = 0

( n-10 ) , (n+3) = 0

n = 10, -3

Negative value of n will not be taken as answer , so n = 10 is the answer..

Thank you