Question

If np3=336,find the value of n??

Answer 1

The value of n is 6.

Step-by-step explanation:

We have,

$^{n}P_3 =336$

To find, the value of n = ?

∴ $^{n}P_3 =336$

⇒ $\dfrac{n!}{(n-3)!} =336$

[ ∵ $^{n}P_r =\dfrac{n!}{(n-r)!}$]

⇒ $\dfrac{n!}{(n-3)!} =336$

⇒$\dfrac{n(n-1)(n-2)(n-3)!}{(n-3)!} =336$

⇒$n(n-1)(n-2)=336$

⇒$n(n-1)(n-2)=$ 6 × 7 × 8

∴ n = 6

Hence, the value of n is 6.