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

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(A) The expression $^nC_r + ^nC_{r-1}$ follows the Pascal's identity for combinations.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$^nC_r +

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$^{n+1}C_r$

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(A) The expression $^nC_r + ^nC_{r-1}$ follows the Pascal's identity for combinations.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$^nC_r + ^nC_{r-1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}$Taking common factors out:$= \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$.

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

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