A) \[\frac{1}{102}\]
B) \[\frac{1}{112}\]
C) \[\frac{1}{122}\]
D) \[\frac{1}{132}\]
Correct Answer: D
Solution :
Probability of sitting all girls together \[=\frac{(6+1)!6!}{12!}\] \[=\frac{7!6!}{12.11.10.9.8.7!}\] \[=\frac{1.2.3.4.5.6}{12.11.10.9.8}\] \[=\frac{1}{132}\]You need to login to perform this action.
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