Question

Prove that c(n r)+c(n r-1)=c(n+1 r)

Answer 1

Answer:

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Answer 2

Answer with Step-by-step explanation:

LHS

$nC_r+nC_{r-1}$

Formula:

$nC_r=\frac{n!}{r!(n-r)!}$

$n!=n(n-1)(n-2)...2\times 1$

Using the formula

$\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$

$\frac{n!}{r(r-1)!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)(n-r)!}$

$\frac{n!}{(r-1)!(n-r)!}(\frac{1}{r}+\frac{1}{n-r+1})$

$\frac{n!}{(r-1)!(n-r)!}(\frac{n-r+1+r}{r(n-r+1)}$

$\frac{n!}{(r-1)!(n-r)!}(\frac{n+1}{r(n-r+1)}$

$\frac{(n+1)!}{r!(n-r+1)!}$

$^{n+1}C_r$

LHS=RHS

Hence, proved.

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