A) \[(9,10)\]
B) \[(10,9)\]
C) \[(7,10)\]
D) \[(10,7)\]
Correct Answer: A
Solution :
[a] \[\frac{2}{9!}+\frac{2}{3!7!}+\frac{1}{5!5!}\] \[=\frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{3!7!}+\frac{1}{9!1!}\] \[=\frac{1}{10!}{{\{}^{10}}{{C}_{1}}+{{\,}^{10}}{{C}_{3}}+{{\,}^{10}}{{C}_{5}}+{{\,}^{10}}{{C}_{7}}+{{\,}^{10}}{{C}_{9}}\}\] \[=\frac{1}{10!}({{2}^{10-1}})=\frac{{{2}^{9}}}{10!}=\frac{{{2}^{a}}}{b!}\] (given) \[\Rightarrow a=9,b=10\]
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