Question

Find:

Answer 1

Answer: $^nC_r=\dfrac{n(n-1)}{2}$

$^nP_{n-2}=n(n-1)$

Step-by-step explanation:

Since we have given that

C(n,n-2)

As we know that

$^nC_r=\dfrac{n!}{(n-r)!r!)}\\\\so,\\\\^nC_{n-2}=\dfrac{n!}{(n-)n-2)!2!}\\\\^nC_{n-2}=\dfrac{n!}{2!(n-2)!}\\\\^nC_{n-2}=\dfrac{n(n-1)(n-2)!}{2!(n-2)!}\\\\^nC_{n-2}=\dfrac{n(n-1)}{2}$

Similarly,

$^nP_{n-2}=\dfrac{n!}{(n-2)!}\\\\^nP_{n-2}=\dfrac{n(n-1)(n-2)!}{(n-2)!}\\\\^nP_{n-2}=n(n-1)$

Hence, $^nC_r=\dfrac{n(n-1)}{2}$

$^nP_{n-2}=n(n-1)$