isi-entrance 2012 Q29

isi-entrance · India · solved Combinations & Selection Combinatorial Identity or Bijection Proof

Using AM-GM inequality, find a lower bound for $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right)$, and determine which of the following holds:
(A) $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right) < \left(\dfrac{2^{10}}{11}\right)^{11}$
(B) $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right) = \left(\dfrac{2^{10}}{11}\right)^{11}$
(C) $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right) > \left(\dfrac{2^{10}}{11}\right)^{11}$
(D) None of the above

By AM-GM: $\frac{{}^{10}C_0 + {}^{10}C_1 + \cdots + {}^{10}C_{10}}{11} > \left({}^{10}C_0 \cdot {}^{10}C_1 \cdots {}^{10}C_{10}\right)^{1/11}$, so $\left(\frac{2^{10}}{11}\right)^{11} > {}^{10}C_0 \cdot {}^{10}C_1 \cdots {}^{10}C_{10}$. (A)

Using AM-GM inequality, find a lower bound for $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right)$, and determine which of the following holds:

(A) $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right) < \left(\dfrac{2^{10}}{11}\right)^{11}$

(B) $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right) = \left(\dfrac{2^{10}}{11}\right)^{11}$

(C) $\left({}^{10}C_0\right)\left({}^{10}C_1\right)\cdots\left({}^{10}C_{10}\right) > \left(\dfrac{2^{10}}{11}\right)^{11}$

(D) None of the above

Paper Questions

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