If P_{n} denotes the product of the binomial | vedclass

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(D) The binomial coefficients in the expansion of $(1+x)^{n}$ are $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$.Thus,$P_{n} = \prod_{k=0}^{n} \bin

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$\frac{(n+1)^{n+1}}{(n+1)!}$

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(D) The binomial coefficients in the expansion of $(1+x)^{n}$ are $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$.Thus,$P_{n} = \prod_{k=0}^{n} \binom{n}{k} = \prod_{k=0}^{n} \frac{n!}{k!(n-k)!} = \frac{(n!)^{n+1}}{\prod_{k=0}^{n} k! (n-k)!} = \frac{(n!)^{n+1}}{(n!)^2} = \frac{(n!)^{n-1}}{1} = \frac{(n!)^{n+1}}{(n!)^2} = \frac{(n!)^{n+1}}{(n!)^2}$.Actually,$P_{n} = \frac{(n!)^{n+1}}{(0! 1! \dots n!)^2}$.Using the property $\prod_{k=0}^{n} \binom{n}{k} = \frac{(n!)^{n+1}}{(0! 1! \dots n!)^2}$,we find the ratio:$\frac{P_{n+1}}{P_n} = \frac{(n+1)!^{n+2}}{\prod_{k=0}^{n+1} (k!)^2} 	imes \frac{\prod_{k=0}^{n} (k!)^2}{(n!)^{n+1}} = \frac{(n+1)^{n+1}}{(n+1)!}$.Therefore,the correct option is $D$.

If $P_{n}$ denotes the product of the binomial coefficients in the expansion of $(1+x)^{n}$,then $\frac{P_{n+1}}{P_n}=$

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