A) \[7\ !\ {{\times }^{6}}{{P}_{3}}\]
B) \[7\ !\ {{\times }^{8}}{{P}_{3}}\]
C) \[7\ !\ \times 3\ !\]
D) \[\frac{10\ !}{3\ !\ 7\ !}\]
Correct Answer: B
Solution :
Seven boys can be seated in a row in \[7\ !\] ways. Hence the total no. of arrangement such that no two girls seated together\[=7\ !\ {{\times }^{8}}{{P}_{3}}\].
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