5 girls and 6 boys are seated on rectangular table with 6 chairs on either side of longer edge. a.What is the total no. of ways the group could be seated?(sides are indistinguishable) b.What is the no. of ways they can be seated so that all 5 girls were sitting on same side?

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A) There are no restrictions in this case. We have 12 chairs and 11 people. We can select an empty chair in 6 ways (because the two sides are indistinguishable). In the remaining 11 chairs, 11 people can be accommodated in 11! ways. So, in total 6*11! ways.

B) If all the 5 girls sit on the side of the empty chair, then we can select an empty chair in $$^6C_1$$ ways, boys can be arranged in 6! ways and girls can be arranged in 5! ways => $$^6C_1*6!*5!$$.
If 5 boys sit on the side of the blank chair, then one boy must sit on the side of the girls. We can select 1 boy in $$^6C_1$$ ways and one blank chair in $$^6C_1$$ ways. Now, the boys can be arranged in 5! ways and the (5 girls + 1 boy) can be arranged in 6! ways => $$^6C_1 * ^6C_1 * 6! * 5!$$.
In total, 42*6!*5!.