If binom{n+1}{3} = 2 times binom{n}{2},then n | vedclass

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(C) Given the equation: $\binom{n+1}{3} = 2 	imes \binom{n}{2}$Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(n+1)!}{3!(n-2)!

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$5$

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(C) Given the equation: $\binom{n+1}{3} = 2 	imes \binom{n}{2}$Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have:$\frac{(n+1)!}{3!(n-2)!} = 2 	imes \frac{n!}{2!(n-2)!}$Since $(n+1)! = (n+1) 	imes n!$ and $3! = 6$,$2! = 2$:$\frac{(n+1) 	imes n!}{6 	imes (n-2)!} = 2 	imes \frac{n!}{2 	imes (n-2)!}$Canceling $n!$ and $(n-2)!$ from both sides:$\frac{n+1}{6} = \frac{2}{2}$$\frac{n+1}{6} = 1$$n+1 = 6$$n = 5$

If $\binom{n+1}{3} = 2 \times \binom{n}{2}$,then $n = \dots$

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