Answer:
[tex]n = 11[/tex]
Step-by-step explanation:
Given
[tex]^nP_2 = 110[/tex]
Required
Solve for n
[tex]^nP_2 = 110[/tex]
Start by applying permutation formula:
[tex]\frac{n!}{(n-2)!} = 110[/tex]
Expand the numerator
[tex]\frac{n*(n-1) * (n-2)!}{(n-2)!} = 110[/tex]
(n-2)! cancels out
[tex]n*(n-1) = 110[/tex]
Expand
[tex]n^2 - n = 110[/tex]
Subtract 110 from both sides
[tex]n^2 - n -110= 110-110[/tex]
[tex]n^2 - n -110= 0[/tex]
Expand:
[tex]n^2 - 11n +10n-110= 0[/tex]
Factorize:
[tex]n(n-11) +10(n-11)= 0[/tex]
[tex](n+10)(n-11) = 0[/tex]
This implies that:
[tex]n + 10 = 0\ or\ n - 11 = 0[/tex]
[tex]n = -10\ or\ n = 11[/tex]
But n can't be negative.
So:
[tex]n = 11[/tex]
