Question

Plz ans it ......urgent
find n...???

Answer 1

Given np4 = 12 * np2

$= \ \textgreater \ \frac{n!}{(n - 4)!} = 12 * \frac{n!}{(n - 2)!}$

$= \ \textgreater \ \frac{n(n - 1)(n - 2)(n - 3)(n - 4)!}{(n - 4)!} = 12 * \frac{n(n - 1)(n - 2)}{(n - 2)!}$

$= \ \textgreater \ n(n - 1)(n - 2)(n - 3) = 12n(n - 1)$

$= \ \textgreater \ (n - 2)(n - 3) = 12$

$= \ \textgreater \ n^2 - 3n - 2n + 6 = 12$

$= \ \textgreater \ n^2 - 5n - 6 = 0$

= > (n + 1)(n - 6) = 0

= > n = -1, n = 6.


Therefore the value of n = 6.


Hope this helps!