If ^nC_3 + ^nC_4 ^{n+1}C_3,then | vedclass

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(A) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.Applying this to the given inequality: $^nC_3 + ^nC_4 = ^{n+1}C_4$.So,the inequality

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$n > 6$

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(A) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.Applying this to the given inequality: $^nC_3 + ^nC_4 = ^{n+1}C_4$.So,the inequality becomes: $^{n+1}C_4 > ^{n+1}C_3$.Expanding the combinations: $\frac{(n+1)!}{4!(n-3)!} > \frac{(n+1)!}{3!(n-2)!}$.Dividing both sides by $(n+1)!$ and simplifying the factorials:$\frac{1}{4 	imes 3!(n-3)!} > \frac{1}{3!(n-2)(n-3)!}$.$\frac{1}{4} > \frac{1}{n-2}$.Since $n > 3$ for the combinations to be defined,$n-2$ is positive,so we can cross-multiply:$n - 2 > 4$.$n > 6$.

If $^nC_3 + ^nC_4 > ^{n+1}C_3$,then

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