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

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(A) Given the inequality: $^nC_3 + ^nC_4 > ^{n+1}C_3$Using the Pascal's identity formula,$^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$,we get:$^{n+1}C_4 > ^{n+1

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$n > 6$

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(A) Given the inequality: $^nC_3 + ^nC_4 > ^{n+1}C_3$Using the Pascal's identity formula,$^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$,we get:$^{n+1}C_4 > ^{n+1}C_3$Dividing both sides by $^{n+1}C_3$ (assuming $n+1 \ge 4$):$\frac{^{n+1}C_4}{^{n+1}C_3} > 1$Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have:$\frac{(n+1)-4+1}{4} > 1$$\frac{n-2}{4} > 1$$n-2 > 4$$n > 6$

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

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