In a group of boys, two boys are brothers and six more boys are present in the group. In how many ways can they sit if the brothers are not to sit along with each other?

2 \times 6!

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^{7} P_{2} \times 6!

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^{7} C_{2} \times 6!

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None of these

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Solution

The correct option is B $^{7} P_{2} \times 6!$
Seating arrangement can be shown as
$\times B_{1} \times B_{2} \times B_{3} \times B_{4} \times B_{5} \times B_{6} \times$ , where $\times$ is two boys who are brother.
Let first six boys sit, which can be done in 6! ways. Once they have been seated,
the two brothers can be made to occupy seats in between or in extreme (i.e. on crosses) in $^{7} P_{2}$ ways.
Hence, required number of ways is $^{7} P_{2} \times 6!$ .

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