The problem is not defined well enough to know what the answer should be. For one thing, the problem says the people are all supposed to get across in 2 round trips. If there are in fact 2 round trips, then somebody ends up on the original side of the river -- unless the boat can go back across the second time without carrying any people. Strange!
On the surface, there are only 2 ways the 10 people can get across:
(1) 6 people cross, 1 returns, 5 people cross; or
(2) 5 people cross, 1 returns, 6 people cross
The answer choices are far larger than "2"; so we must assume that we are in fact choosing among distinguishable people, so "a choose b"comes into play.
I see nothing wrong with the work I have done so far on the problem, except that I have only considered the scenario when 6 people make the trip across the first time. My work shows
(10C6) * (6C1) = 210*6 = 1260
That is, I am choosing 6 of the 10 people to cross first, and then I am choosing 1 of the 6 to return. I feel that is a reasonable way to interpret the problem; however, this answer is already larger than any of the answer choices. So it must be that we aren't supposed to interpret the problem this way.
If you only consider the two cases where you choose either 6 or 5 people go across the first time, without figuring in the choice of the one person who returns, you get one of the given answer choices.
It seems to me that if we are considering which of the people go across the first time, then we should also consider which person returns to get the other people. In that case, the correct answer to the problem would be
(10C6)*(6C1) + (10C5)*(5C1)
So, again, the problem is not presented well enough for us to know how to calculate the answer; and the apparently correct answer uses an interpretation of the problem which is inconsistent, in using "a choose b" for the first trip across but not for the return trip.
