Question

nP5 = 20 nP3, find nC6

Answer 1

See the attach file for ans

Answer 2

Answer:

$^nC_6 =28$

Step-by-step explanation:

Given : $^nP_5 = 20 ^nP_3$

To Find: $^nC_6$

Solution:

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

$^nP_5 = 20 ^nP_3$

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

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

$1 = 20  \times \frac{1}{(n-3)(n-4)}$

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

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

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

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

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

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

$n=-1,8$

Since number of terms cannot be negative so, n =8

Now Formula of combination: $^nC_r=\frac{n!}{r!(n-r)!}$

Since we are supposed to find : $^nC_6$

Since n = 8

So we need to find : $^8C_6$

So, $^8C_6=\frac{8!}{6!(8-6)!}$

$^8C_6=\frac{8!}{6!(2)!}$

$^8C_6=\frac{8 \times 7 \times 6!}{6!(2)!}$

$^8C_6=\frac{8 \times 7 }{2}$

$^8C_6=28$

Hence $^nC_6 =28$