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Q. A class is composed of two brothers and six other boys. In how many ways can all the boys be seated at a round table so that the two brothers are not seated besides each others ?

Permutations and Combinations

Solution:

Total no. of ways $=\left(6-1\right)\,! \times\,{}^{6}p_{2}$
$=5!\times 30=120\times 30$
$=3600$