Question

Solve for n in each of the following equations:
(a)
n^ Pbase2=110​

Answer 1

`n = 11`

Explanation:

Given

`^nP_2 = 110`

Required

Solve for n

`^nP_2 = 110`

Start by applying permutation formula:

`(n!)/((n-2)!) = 110`

Expand the numerator

`(n*(n-1) * (n-2)!)/((n-2)!) = 110`

(n-2)! cancels out

`n*(n-1) = 110`

Expand

`n^2 - n = 110`

Subtract 110 from both sides

`n^2 - n -110= 110-110`

`n^2 - n -110= 0`

Expand:

`n^2 - 11n +10n-110= 0`

Factorize:

`n(n-11) +10(n-11)= 0`

`(n+10)(n-11) = 0`

This implies that:

`n + 10 = 0\ or\ n - 11 = 0`

`n = -10\ or\ n = 11`

But n can't be negative.

So:

`n = 11`