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A class is composed of 2 brothers and 6 other boys. In how many ways can all the boys be
seated at a round table so that the two brothers are not seated together?

a. 3000

b.3600

c. 2050

d. 2600

3 Answers

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1 votes
4320
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as in given question , 2 brother cant sit together so there are 6 place in which they can sit in round table ,

so $\binom{6}{2}$*2!*(6-1)! = 3600 ans
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There are 2 brothers and 6 different students. Firstly, the 6 students can be seated at a round table in (6-1)!=5! ways. Name the boys as B1, B2, B3, B4, B5, B6 in clockwise or anticlockwise order. Assume that between every 2 boys(say B1-B2 or B2-B3, or B3-B4, etc), there is a vacant seat which can be occupied by either of the brothers so that they don't seat next to each other. Following this, the total no of seats=6.

We can now select 2 seats out of 6 in 6C2 ways and both the brothers can arrange themselves in 2! ways.

Final answer=5!*6C2*2!=3600

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