If binom{n}{3} + binom{n}{4} binom{n+1}{3}, | vedclass

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(A) Using the Pascal's identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$,we have:$\binom{n}{3} + \binom{n}{4} = \binom{n+1}{4}$Given the inequ

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$n > 6$

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(A) Using the Pascal's identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$,we have:$\binom{n}{3} + \binom{n}{4} = \binom{n+1}{4}$Given the inequality:$\binom{n+1}{4} > \binom{n+1}{3}$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)!}$$\frac{1}{4} > \frac{1}{n-2}$Since $n-2 > 0$ (as $n \ge 4$ for the combinations to be defined),$n-2 > 4$$n > 6$

If $\binom{n}{3} + \binom{n}{4} > \binom{n+1}{3}$,then:

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