A) \[\frac{{{k}^{2}}}{432}\]
B) \[\frac{k(k-1)}{432}\]
C) \[\frac{(k-1)(k-2)}{432}\]
D) \[\frac{k(k-1)(k-2)}{432}\]
Correct Answer: C
Solution :
\[n(S)={{6}^{3}}=6\times 6\times 6=216\] n(S)= Coefficient of \[{{x}^{k}}\] in\[{{(x+{{x}^{2}}+...+{{x}^{6}})}^{3}}\] = Coefficient of \[{{x}^{k-3}}\]in\[{{\left( \frac{1-{{x}^{6}}}{1-x} \right)}^{3}}\] = Coefficient of \[{{x}^{k-3}}\] in \[{{(1-{{x}^{6}})}^{3}}{{(1-x)}^{-3}}\] = Coefficient of \[{{x}^{k-3}}\]in \[{{(1-x)}^{-3}}\] \[(\because 3\le k\le 8)\] \[{{=}^{k-3+3-1}}{{C}_{3-1}}\] \[{{=}^{k-1}}{{C}_{2}}=\frac{(k-1)(k-2)}{2}\] Hence, the probability of the required event \[=\frac{(k-1)(k-2)}{2\times 216}=\frac{(k-1)(k-2)}{432}\] Solutions for Q. No. 16 to 17.
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