Answer:
[tex]n = 8[/tex]
Step-by-step explanation:
Given
[tex]3(^nP_2 + 24) = ^{2n}P_2[/tex]
Required
Find n
To do this, we simply apply permutations formula
[tex]nP_r = \frac{n!}{(n -r)!}[/tex]
So, we have:
[tex]3 * [\frac{n!}{(n -2)!} + 24] = \frac{2n!}{(2n -2)!}[/tex]
Expand
[tex]3 * [\frac{n * (n - 1) * (n - 2)!}{(n -2)!} + 24] = \frac{2n * (2n - 1) * (2n - 2)}{2n - 2}[/tex]
[tex]3 * [n * (n - 1) + 24] = 2n * (2n - 1)[/tex]
[tex]3 * [n^2 - n + 24] = 4n^2 - 2n[/tex]
Open bracket
[tex]3n^2 - 3n + 72 = 4n^2 - 2n[/tex]
Collect like terms
[tex]3n^2 - 4n^2- 3n+2n + 72 = 0[/tex]
[tex]-n^2- n + 72 = 0[/tex]
Expand
[tex]-n^2 -9n + 8n + 72 = 0[/tex]
Factorize
[tex]-n(n +9) - 8(n + 9) = 0[/tex]
Factor out n + 9
[tex](-n -8)(n + 9) = 0[/tex]
Split
[tex](-n -8)= 0 \ or\ (n + 9) = 0[/tex]
Solve for n
[tex]n =8\ or\ n = -9[/tex]
The positive value is [tex]n = 8[/tex]
