Answer:
[tex]n = 5[/tex]
Step-by-step explanation:
Given
[tex]^nP_2 = 20[/tex]
Required
Find n
To do this, we apply permutation formula.
[tex]^nP_r = \frac{n!}{(n-r)!}[/tex]
So, we have:
[tex]\frac{n!}{(n-2)!} = 20[/tex]
Expand the numerator
[tex]\frac{n(n-1)(n-2)!}{(n-2)!} = 20[/tex]
Divide by (n-2)!
[tex]n(n-1)= 20[/tex]
Expand
[tex]n^2 - n = 20[/tex]
[tex]n^2 - n - 20 = 0[/tex]
Expand
[tex]n^2 +4n -5n - 20 = 0[/tex]
Factorize
[tex]n(n+4)-5(n+4)= 0[/tex]
[tex](n-5)(n+4) = 0[/tex]
[tex]n - 5 = 0[/tex] or [tex]n + 4 = 0[/tex]
[tex]n = 5[/tex] or [tex]n = -4[/tex]
Since n cannot be negative, then:
[tex]n = 5[/tex]
