Question

Find:

a. C(n,n-2)____________________________

b. P(n,n-2)____________________________

Answer 1

Answer:
`^nC_r=(n(n-1))/(2)`

`^nP_(n-2)=n(n-1)`

Explanation:

Since we have given that

C(n,n-2)

As we know that

`^nC_r=(n!)/((n-r)!r!))\\\\so,\\\\^nC_(n-2)=(n!)/((n-)n-2)!2!)\\\\^nC_(n-2)=(n!)/(2!(n-2)!)\\\\^nC_(n-2)=(n(n-1)(n-2)!)/(2!(n-2)!)\\\\^nC_(n-2)=(n(n-1))/(2)`

Similarly,

`^nP_(n-2)=(n!)/((n-2)!)\\\\^nP_(n-2)=(n(n-1)(n-2)!)/((n-2)!)\\\\^nP_(n-2)=n(n-1)`

Hence,
`^nC_r=(n(n-1))/(2)`

`^nP_(n-2)=n(n-1)`