Question

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 ?

• A. $720$
• B. $1440$
• C. $3600$
• D. $4320$

Answer 1

The Correct Option is C

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

Answer 2

Total arrangements of 8 boys at a round table:
Fix one boy's position (to eliminate rotational symmetry).
Arrange the remaining 7 boys.
\(7! = 5040\)

Arrangements where the two brothers sit together:
Treat the two brothers as a single "block", so we have 7 units to arrange (6 other boys + 1 block).
Fix one unit's position, then arrange the remaining 6 units.
\(6! = 720\)
The two brothers within the block can swap positions.
\(720 \times 2 = 1440\)

Arrangements where the two brothers do not sit together:
Subtract the arrangements where they sit together from the total arrangements.
\(5040 - 1440 = 3600\)

So, the correct option is (C): 3600