Question

prove that n! /(n-r)! r! +n! /(r-1)!(n-r+1)!=(n+1)!/r!(n-r+1)!​

Answer 1

Okay. I prove it even I didn't learn the combinations!!! See below!

Actually we're going to prove that,

$\boxed{^nC_r+ \ ^nC_{r-1}=\ ^{n+1}C_r}$

So let's start!!!

LHS

$^nC_r+\ ^nC_{r-1} \\ \\ \frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!} \\ \\ \frac{n(n-1)(n-2)...(n-r+1)(n-r)!}{r!(n-r)!}+\frac{n(n-1)(n-2)...(n-r+2)(n-r+1)!}{(r-1)!(n-r+1)!} \\ \\ \frac{n(n-1)(n-2)...(n-r+1)}{r!}+\frac{n(n-1)(n-2)...(n-r+2)}{(r-1)!} \\ \\ \frac{n(n-1)(n-2)...(n-r+1)}{r!}+\frac{rn(n-1)(n-2)...(n-r+2)}{r(r-1)!} \\ \\ \frac{n(n-1)(n-2)...(n-r+1)}{r!}+\frac{rn(n-1)(n-2)...(n-r+2)}{r!} \\ \\ \frac{n(n-1)(n-2)...(n-r+1)+rn(n-1)(n-2)...(n-r+2)}{r!}$

$\frac{n(n-1)(n-2)...(n-r+1)+rn(n-1)(n-2)...(n-r+2)}{r!} \\ \\ \frac{[n(n-1)(n-2)...(n-r+2)]\ n-r+1+r}{r!} \\ \\ \frac{[n(n-1)(n-2)...(n-r+2)]\ n+1}{r!} \\ \\ \frac{(n+1)n(n-1)(n-2)...(n-r+2)(n-r+1)!}{r!(n-r+1)!} \\ \\ \frac{(n+1)!}{r(n-r+1)!} \\ \\ ^{n+1}C_r$

RHS

Hence proved!!!

Pls ask me if you have any doubt on my answer.

Thank you...

Answer 2

your answer attached in the photo