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Question

Prove that:
(i) n!(nr)!
= n(n-1)(n-2)...(n-(r-1))
(ii) n!(nr)!r!+n!(nr+1)!(r1)!
= (n+1)!r!(nr+1)!

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Solution

(i) We have,
LHS = n!(nr)!
= n(n1)(n2)(n3)...(nr+2)(nr+1)(nr)!(nr)!
= n(n1)(n2)(n3)...(nr+2)(nr+1)
= n(n1)(n2)(n3)...((n(r2))(n(r1))
= n(n1)(n2)(n3)...(n(r1))
= RHS
LHS =RHS
(ii) To prove n!(nr)!r!+n!(nr+1)!(r1)!=(n+1)!r!(nr+1)!
LHS=n!(nr)!r!+n!(nr+1)!(r1)!
=n!(nr)! (r) [(r1)]!+n!(nr+1)[(nr)!](r1)!
= n!(nr)!×(r1)![1r+1nr+1]= n!(nr)!×(r1)![nr+1+rr(nr+1)]
= n!(nr)!×(r1)![n+1r(nr+1)]= (n+1)×n!(nr+1)×(nr)!×r×(r1)!
= (n+1)!(nr+1)!×r!=(n+1)!r!(nr+1)!
=RHS
LHS = RHS


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