3 on the bow side can be arranged in 4P3 ways,
Same is for the other 3 on the other side.
The remaining 2 can be arranged in 2 ways

So total number of arrangements would be
4P3*4P3*2=1152

IMO E

Posted from my mobile device

There are 2 rowers who can sit on either side. They can arrange themselves in 2 ways.

The 4 rowers on the stroke side can be then arranged in 4! ways and those on the bow side can arrange themselves in 4! ways.

Therefore the total number of ways = 2 * 4! * 4! = 2 * 24 * 24 = 1152


Option E

Arun Kumar
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Should it not be 4C3, since the ordering you do later, when multiplying by 3!

My solution is:
Left - 4C3 * 3!
Right - 4C3 * 3!
And the 2 spots (one in each side) can basically be swapped, so 2!

so (4C3*3!)^2 * 2!
=1152 (E)

4P3 includes the ordering, so I did not multiply with 3!
(4C3*3! Is same as 4P3)

Solution:

Since the 3 people must sit on the stroke side, they have 4P3 = 4 x 3 x 2 = 24 ways to choose their seats on the stroke side. Likewise, since another 3 people must sit on the bow side, they have 4P3 = 4 x 3 x 2 = 24 ways to choose their seats on the bow side. For the remaining 2 people who can sit anywhere in the remaining 2 unpicked seats, they have 2! = 2 ways to choose their seats. Therefore, there are a total of 24 x 24 x 2 = 1152 possible sitting arrangements.

Answer: E
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