A) \[{{(7\,!)}^{2}}\]
B) \[7\,!\,\times \,6\,!\]
C) \[{{(6\,!)}^{2}}\]
D) \[7\,!\]
Correct Answer: B
Solution :
Fix up 1 man and the remaining 6 men can be seated in 6! ways. Now no two women are to sit together and as such the 7 women are to be arranged in seven empty seats between two consecutive men and number of arrangement will be 7!. Hence by fundamental theorem the total number of ways = 7! × 6!.
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