Let n∈N, n≥4 & P=∏r=0n nCr, then

P < 2 n − 2 n − 1 n − 1 , P < 2 n n + 1 n + 1 , P < 2 n − 2 n − 1 n

Let n∈N, n≥4 & P=∏r=0n nCr, then

n C 1 n C 2 … … n C n − 1 1 n − 1 < n C 1 + n C 2 + … … + n C n − 1 n − 1 P < 2 n − 2 n − 1 n − 1 Also n C 10 n C 1 n C 2 … … ⋅ n C n − 1 n C n 1 n − 1 < n C 0 + n C 1 + n C 2 + … … + n C n n + 1 P < 2 n n + 1 n + 1

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Let $n \in N, n \geq 4 & P = \prod_{r = 0}^{n}^{n} C_{r}$ , then

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$P > {(\frac{2^{n}}{n + 1})}^{n + 1}$

$P < {(\frac{2^{n} - 2}{n - 1})}^{n - 1}$

$P < {(\frac{2^{n}}{n + 1})}^{n + 1}$

$P < {(\frac{2^{n} - 2}{n - 1})}^{n}$

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} {(^{n} {C_{1}}^{n} C_{2} \dots \dots^{n} C_{n - 1})}^{\frac{1}{n - 1}} \\ < \frac{^{n} C_{1} +^{n} C_{2} + \dots \dots +^{n} C_{n - 1}}{n - 1} P < {(\frac{2^{n} - 2}{n - 1})}^{n - 1} \\ Also {(^{n} {C_{10}}^{n} C_{1}^{n} C_{2} \dots \dots \cdot^{n} {C_{n - 1}}^{n} C_{n})}^{\frac{1}{n - 1}} \\ < \frac{^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots \dots +^{n} C_{n}}{n + 1} P < {(\frac{2^{n}}{n + 1})}^{n + 1} \end{array}$