Question

np5 = 20.np3 then what will be the value of n?

Answer 1

Answer:

n=8

Step-by-step explanation:

Formula of permutation : $^nP_r=\frac{n!}{(n-r)!}$

We are given that $^nP_5 = 20 \times^nP_3$

So, Using formula :

$\frac{n!}{(n-5)!}= 20 \times\frac{n!}{(n-3)!}$

$\frac{n \times (n-1) \times (n-2) \times (n-3) \times (n-4) \times (n-5)!}{(n-5)!}= 20 \times \frac{n \times (n-1) \times (n-2) \times (n-3)!}{(n-3)!}$

$(n-3) \times (n-4) = 20$

$n \times (n-4) -3 \times (n-4) = 20$

$n^2-4n -3n +12 -20 =0$

$n^2-7n -8 =0$

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

$n(n+1)-8(n+1) =0$

$(n+1)(n-8) =0$

$n=-1,8$

Since n cannot be negative

So, n = 8