Question

N
2. (a) If ^nP2= 20, find the value of n.
2​

Answer 1

`n = 5`

Explanation:

Given

`^nP_2 = 20`

Required

Find n

To do this, we apply permutation formula.

`^nP_r = (n!)/((n-r)!)`

So, we have:

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

Expand the numerator

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

Divide by (n-2)!

`n(n-1)= 20`

Expand

`n^2 - n = 20`

`n^2 - n - 20 = 0`

Expand

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

Factorize

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

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

`n - 5 = 0` or
`n + 4 = 0`

`n = 5` or
`n = -4`

Since n cannot be negative, then:

`n = 5`