Answer: [tex]^nC_r=\dfrac{n(n-1)}{2}[/tex]
[tex]^nP_{n-2}=n(n-1)[/tex]
Step-by-step explanation:
Since we have given that
C(n,n-2)
As we know that
[tex]^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}[/tex]
Similarly,
[tex]^nP_{n-2}=\dfrac{n!}{(n-2)!}\\\\^nP_{n-2}=\dfrac{n(n-1)(n-2)!}{(n-2)!}\\\\^nP_{n-2}=n(n-1)[/tex]
Hence, [tex]^nC_r=\dfrac{n(n-1)}{2}[/tex]
[tex]^nP_{n-2}=n(n-1)[/tex]
