A) \[8\]
B) \[10\]
C) \[12\]
D) \[15\]
Correct Answer: C
Solution :
[c] Let the consecutive vertices be \[{{A}_{1}},{{A}_{2}},{{A}_{3}},.....,{{A}_{n-1}},An.\] 1 st vertex can be selected in n ways. If the first vertex is \[{{A}_{1}},\] then vertices \[{{A}_{2}}\] and \[{{A}_{n}}\] cannot be selected. So, now we have \[(n-3)\] vertices available for four remaining vertices of pentagon. These four vertices are not consecutive. Then we have to find the number of integral solutions of the equation \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}=n-7;\] where \[{{x}_{1}},{{x}_{5}}\ge 0\] and \[{{x}_{2}},{{x}_{3}},{{x}_{4}}\ge 1\] or \[{{x}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}+{{x}_{5}}=n-10;\] where \[{{x}_{1}},{{x}_{5}}\ge 0\]and \[{{y}_{2}},{{y}_{3}},{{y}_{4}}\ge 0\] So, number of non-negative integral solutions \[{{=}^{n-10+5-1}}{{C}_{5-1}}{{=}^{n-6}}{{C}_{4}}\] So, total number of ways \[=\frac{^{n-6}{{C}_{4}}.n}{5}\] \[\Rightarrow \,\,\,\,\,\,\frac{^{n-6}{{C}_{4}}.n}{5}=36\] (Given) \[\therefore \,\,\,\,\,n=12\]
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