Three small cruise ships, each carrying 10 passengers, will dock

@PORT A: The gifts have to be given to 2 out of 10 members who can be selected in 10C2 ways = 45 ways
Please Note: Here the arrangement doesn't come into picture because both the gift values are same. If the gift values were different then it would have required the arrangement of Gifts between the 2 selected candidates out of 10

@PORT B:The gift has to be given to 1 out of 10 members who can be selected in 10C1 ways = 10 ways

@PORT C:The gift has to be given to 1 out of 10 members who can be selected in 10C1 ways = 10 ways

Since there are 3 different ships carrying different sets of 10 people each so the group of 10 people taking gifts at different ports can also be arranged and the arrangements of three Ships at three ports = 3! = 6

Total Ways to give gifts on all three ports = 45*10*10*6 = 27000 ways

Answer: Option E

MANHATTAN GMAT OFFICIAL SOLUTION:

This problem features four separate decisions. In other words, you must answer four separate questions:

(1) Which ship docks at which port?
(2) Who are the two people who receive the two gift certificates at Port A?
(3) Who receives the one gift certificate at Port B?
(4) Who receives the one gift certificate at Port C?

The first of these decisions is a permutation, because "order matters." In other words, switching any of the port assignments results in a new arrangement.

Since the gift certificates are worth different amounts at different ports, the specific way in which we assign ships to ports matters.

Thus, we calculate the number of permutations of the 3 ships over the 3 ports as 3! = 6.

The second decision is a combination, because "order does not matter." You do not care about the sequence in which you choose the two winners. You only care about who has won. Since there are 2 winners and 10 - 2 = 8 losers out of a pool of 10 contestants at Port A, we can write the number of combinations as 10!/(2!8!) = 45.

Finally, the third and fourth decisions are very simple combinations. You have 10 choices for who receives the certificate at Port B, and separately you have 10 choices for who receives the certificate at Port C.

The decisions are sequential, and they are made independently. As a result, the numbers of possibilities are multiplied at the end of the problem. The total number of ways the gift certificates can be given out is therefore

3!*10!/(2!8!)*10*10 = 27,000.

Answer: E.

E.
we have 3 ships with different people in all 3 ships(no 2 people can be same). So, if we have 3 batches of people, eg B1, B2, B3 and each has 10 people. in that way these batches could dock on any 3 ports out of A, B and C as given in Q.
So if ABC is the sequence of ports, that way we will have B1 B2 B3 docking respectively at A B and C.

Now these B1 B2 and B3 could shuffle positions b/w A, B and C port. (it could be B1B2B3 or B1B3B2 or B2B1B3 etc.)
in this way, we will have 6 different ways of arranging these 3 batches of people.
now if any batch comes on port A = we have 10C2 => 45, on port B = 10C1 =>10 and on port C = 10C1 => 10. Thus 45*10*10 => 4500.
Thus we will have 4500*6 = 27000 ways of giving out certificates.
(since all these people will be different, we need to keep them in batches as shown above. Also, since any Batch could dock on Any port that's why B1,B2, B3 could dock on any of A or B or C port. Had these batches been constrained or fixed at particular ports in Question, we then wouldnt require the multiplication of 6.)

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