Prove that:
(i) n!(n−r)!
= n(n-1)(n-2)...(n-(r-1))
(ii) n!(n−r)!r!+n!(n−r+1)!(r−1)!
= (n+1)!r!(n−r+1)!
(i) We have,
LHS = n!(n−r)!
= n(n−1)(n−2)(n−3)...(n−r+2)(n−r+1)(n−r)!(n−r)!
= n(n−1)(n−2)(n−3)...(n−r+2)(n−r+1)
= n(n−1)(n−2)(n−3)...((n−(r−2))(n−(r−1))
= n(n−1)(n−2)(n−3)...(n−(r−1))
= RHS
∴ LHS =RHS
(ii) To prove n!(n−r)!r!+n!(n−r+1)!(r−1)!=(n+1)!r!(n−r+1)!
LHS=n!(n−r)!r!+n!(n−r+1)!(r−1)!
=n!(n−r)! (r) [(r−1)]!+n!(n−r+1)[(n−r)!](r−1)!
= n!(n−r)!×(r−1)![1r+1n−r+1]= n!(n−r)!×(r−1)![n−r+1+rr(n−r+1)]
= n!(n−r)!×(r−1)![n+1r(n−r+1)]= (n+1)×n!(n−r+1)×(n−r)!×r×(r−1)!
= (n+1)!(n−r+1)!×r!=(n+1)!r!(n−r+1)!
=RHS
∴ LHS = RHS