^{n-1}C_3 + ^{n-1}C_4 ^nC_3,then the value | vedclass

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(C) Using the Pascal's identity,$^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r$,we have:$^{n-1}C_3 + ^{n-1}C_4 = ^nC_4$.Given the inequality: $^{n-1}C_3 + ^{n-1}C

Vedclass · Accepted Answer

$> 7$

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(C) Using the Pascal's identity,$^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r$,we have:$^{n-1}C_3 + ^{n-1}C_4 = ^nC_4$.Given the inequality: $^{n-1}C_3 + ^{n-1}C_4 > ^nC_3$.Substituting the identity: $^nC_4 > ^nC_3$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get:$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$.Dividing both sides by $n!$ and simplifying:$\frac{1}{4 \cdot 3!(n-4)!} > \frac{1}{3!(n-3)(n-4)!}$.$\frac{1}{4} > \frac{1}{n-3}$.Since $n-3 > 0$ (as $n \ge 4$ for combinations to be defined),we have:$n-3 > 4 \Rightarrow n > 7$.

$^{n-1}C_3 + ^{n-1}C_4 > ^nC_3$,then the value of $n$ is

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