^nC_r + ^nC_{r-1} is equal to | vedclass

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Question

(A) The expression $^nC_r + ^nC_{r-1}$ is a standard identity in combinatorics known as Pascal's Identity.Using the formula $^nC_r = \frac{n!}{r!(n-r)

Vedclass · Accepted Answer

$^{n+1}C_r$

Vedclass · Answer

(A) The expression $^nC_r + ^nC_{r-1}$ is a standard identity in combinatorics known as Pascal's Identity.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$^nC_r + ^nC_{r-1} = \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)!} \left( \frac{1}{r} + \frac{1}{n-r+1} \right)$$= \frac{n!}{(r-1)!(n-r)!} \left( \frac{n-r+1+r}{r(n-r+1)} \right)$$= \frac{n!}{(r-1)!(n-r)!} \left( \frac{n+1}{r(n-r+1)} \right)$$= \frac{(n+1)n!}{r(r-1)! (n-r+1)(n-r)!}$$= \frac{(n+1)!}{r!(n+1-r)!} = ^{n+1}C_r$Thus,the correct option is $A$.

$^nC_r + ^nC_{r-1}$ is equal to

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