Problem Solving

Ten people want to cross a river by a boat which can carry a maximum of 6 people at a time. If the boat is to be used for two round trips then in how many ways can these people cross the river?

A) 420
B) 504
C) 462
D) 924
E) 1008

IMO D

There are 2 ways in which the boat can take 10 people accorss the river

1. first trip - 4 ppl, second trip - 4 ppl
2. first trip - 6 ppl, second trip - 6 ppl
3. first trip - 5 ppl, second trip - 5 ppl
4. first trip - 5 ppl, second trip - 5 ppl

for first 2, we have 10C6 and 10C4 = 420
for next 2, we have 10C5 and 10C5 = 504
adding up, we get 924

krishnasty wrote:IMO D

There are 2 ways in which the boat can take 10 people accorss the river

1. first trip - 4 ppl, second trip - 4 ppl
2. first trip - 6 ppl, second trip - 6 ppl
3. first trip - 5 ppl, second trip - 5 ppl
4. first trip - 5 ppl, second trip - 5 ppl

for first 2, we have 10C6 and 10C4 = 420
for next 2, we have 10C5 and 10C5 = 504
adding up, we get 924

Are we assuming that the boat going to the other side will return on its own or is there a sailor in the boat?

because when I solved this I counted the number of ways in which people can get down on the other side.

say for example-- if 6 people start from Side A to Side B.

they can do that in 10C6 ways .. now any 5 will get down on the other side..

this will be done in 6C5 ways = 6 ways and the person left in the boat will sail back to side A to pick the rest of the people

Am I missing something ?

I have answered the question assuming that the boat has a sailor and the max 6 ppl capacity is exclusive of the sailor.
If this would have been a trick question, you could get multiple combinations which could exceed the max option given. Plus, i am not sure if that would have been a GMAT question.

Still i would like to run my solution through some GMAT expert.

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.

THE OA is D, so I am assuming the boat had a separate driver and the question is only the number of ways the 10 people can cross the river (excluding the driver)

I get 10C6 and 10C4 when the first trip includes 6 people and 4 people respectively
but for 10C5, dont get why we multiply it twice ( doesnt 10C5 already include all possibility of selecting 5 people for the first trip? and remaining 5 people would be 5c5 ?) Obviously I am wrong here, could someone explain?