A) \[{{(10!)}^{2}}\left( \frac{2}{3!2!7!}+\frac{1}{{{(2!)}^{5}}6!} \right)\]
B) \[{{(10!)}^{2}}\left( \frac{1}{3!2!7!}+\frac{1}{{{(2!)}^{4}}6!} \right)\]
C) \[{{(10!)}^{2}}\left( \frac{1}{3!7!}+\frac{1}{{{(2!)}^{5}}6!} \right)\]
D) \[{{(10!)}^{2}}\left( \frac{1}{3!7!}+\frac{1}{{{(2!)}^{5}}(6!)} \right)\]
Correct Answer: B
Solution :
It is possible in two mutually exclusive cases. Case 1.2 children get none, one child gets three and all remaining children get one each. The number of ways\[=\frac{10!}{3!2!7!}.10!\]Case n. 2 children get none, 2 children get 2 each and all remaining children get one each. The number of ways\[=\frac{10!}{{{(2!)}^{4}}6!}.10!\] \[\therefore \]Total number of ways\[={{(10!)}^{2}}\left( \frac{1}{3!2!7}+\frac{1}{{{(2!)}^{4}}6!} \right)\]
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