Question

Find the value of n such that :

nP5 ,= 42n P3 , n > 4​

Answer 1

Answer:

n = 9, 10

Step-by-step explanation:

⇒ nP5 = 42 × nP3 where, n > 4

⇒ nP5 / nP3 = 42

⇒ nP5 = ( n! ) / ( n - 5 )!

⇒ nP3 = ( n! ) / ( n - 3 )!

We know that,

⇒ ( n - 3 )! = ( n - 3 ) ( n - 4 ) ( n - 5 )!

So substituting this we get,

⇒ nP3 = ( n! ) / ( n - 3 ) ( n - 4 ) ( n - 5 )!

Now Dividing nP5 and nP3 we get,

⇒nP5 / nP3 = ( n! ) / ( n - 5 )! ÷ ( n! ) / ( n - 3 ) ( n - 4 ) ( n - 5 ) ! = 42

Cancelling, ( n! ) and ( n-5 )! we get,

⇒ ( n - 3 ) ( n - 4 ) = 42

Let us consider ( n - 3 ) as x. Then ( n - 4 ) is [ ( n - 3 ) - 1 ] = ( x - 1 )

So, we get,

⇒ x ( x - 1 ) = 42

⇒ x² - x = 42

⇒ x² - 7x + 6x - 42 = 0

⇒ x ( x - 7 ) 6 ( x - 7 ) = 0

⇒ ( x - 7 ) ( x + 6 ) = 0

⇒ x = 7, -6

⇒ n - 3 = 7 ; n - 3 = 6

We know that n > 4. Hence n - 3 = -6 is neglected.

⇒ n - 3 = 7

⇒ n = 7 + 3 = 10

And, n - 4 = n - 3 -1 = 10 - 1 = 9

Hence the possible values of n greater than 4 are 9 and 10.

Hope it helped !!