| In how many ways can 5 boys and 5 girls sit in a circle, so that no two boys sit together? |
A) \[5!\,\times \,5!\]
B) \[4!\,\times \,5!\]
C) \[\frac{5!\,\,\times \,\,5!}{2}\]
D) None of these
Correct Answer: B
Solution :
| First we fix the alternate position of the girls. The number of ways in which five girls can be seated around the circle \[=(5-1)!=4!.\] Now, 5 boys can be seated in five vacant place in 51 ways. |
| \[\therefore \] Required number of ways \[=4!\,\times 5!\] |
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