Question

prove nCr + nCr-1 = n+1Cr​

Answer 1

Given :

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

$\sf LHS = ^nC_r + ^nC_{r-1}$

$\sf = \dfrac{n!}{(n-r)!r!} + \dfrac{n!}{(n-(r-1))!(r-1)!}$

$\sf = \dfrac{n!}{(n-r)!r(r - 1) !} + \dfrac{n!}{(n-r + 1)(n - r)!(r-1)!}$

$\sf = \dfrac{n!}{(n-r)!(r - 1) !} \: \bigg[ \dfrac{1}{r} + \dfrac{1}{n - r + 1} \bigg]$

$\sf = \dfrac{n!}{(n-r)!(r - 1) !} \: \bigg[ \dfrac{n - r + 1 + r}{r(n - r + 1)} \bigg]$

$\sf = \dfrac{n!(n + 1)}{ \bigg[(n-r)!(n - r + 1) !\bigg]\bigg[r(r - 1)!\bigg]}$

$\sf = \dfrac{(n + 1)!}{(n + 1 - r) ! \: r!}$

$\sf = \: ^{n+1}C_r \: \: (RHS)$

LHS = RHS

Hence proved !!