Question

Find the value of combination and permutations if np2=72

Answer 1

np2= 72 . then n( n-1)=72 and the value of n = 9. therefore 9p2=72 and 9c2=36

Answer 2

Answer is 9

Step-by-step explanation:

$n_{2}^{p}=72$

$\frac {n!}{(n-2)!} = 72$

$\frac {n \times (n-1) \times (n-2)!}{(n-2)!} = 72$

$n \times (n-1) = 72$

$n^2 - n - 72 = 0$

$n^2 -9n + 8n -72 = 0$

(n-9) =0, (n+8) =0

n = 9,-8

We take (+) Values

So, value of Combination as well as Permutations is 9