If ^nC_r = ^nC_{r-1} and ^nP_r = ^nP_{r+1},th | vedclass

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(A) Given $^nC_r = ^nC_{r-1}$. Using the property $^nC_a = ^nC_b \implies a = b$ or $a + b = n$,we have $r = r-1$ (impossible) or $r + (r-1) = n$,whic

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$3$

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(A) Given $^nC_r = ^nC_{r-1}$. Using the property $^nC_a = ^nC_b \implies a = b$ or $a + b = n$,we have $r = r-1$ (impossible) or $r + (r-1) = n$,which gives $n = 2r - 1$ or $r = \frac{n+1}{2}$.Given $^nP_r = ^nP_{r+1}$. We know $^nP_r = \frac{n!}{(n-r)!}$ and $^nP_{r+1} = \frac{n!}{(n-r-1)!}$.Equating them: $\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!} \implies \frac{1}{n-r} = 1 \implies n-r = 1 \implies r = n-1$.Substitute $r = n-1$ into $r = \frac{n+1}{2}$:$n-1 = \frac{n+1}{2} \implies 2n - 2 = n + 1 \implies n = 3$.

If $^nC_r = ^nC_{r-1}$ and $^nP_r = ^nP_{r+1}$,then the value of $n$ is

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