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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
| 1k views
+1
my ans is 4320.
0
B. 3600
0
could you plz elaborate?
+2
Take 1 person from 6 and fix him and 5 others can arranged in -- 5! ways=120

there are 6 places left in which 2 brothers can sit

so they can choose any 2 places from 6 - 6C2 ways=15

2 brothers can arrange themselves in 2! ways=15*2=30

total ways=120*30=3600

+1 vote
4320
by (259 points)
+2

Total ways to arrange 8 boys on a circular table would be: (8-1)!​​​​=5040

Ways to arrange the brothers are always together would be: ((7-1)! * 2!)=1440

So total ways that brothers are not allowed to be seated together= 5040-1440=3600

​​​​

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
by Active (3.8k points)
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.

by Active (2k points)

+1 vote