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

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Three small cruise ships, each carrying 10 passengers, will dock  [#permalink]

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New post 07 Jul 2015, 03:04
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Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at Port B, and the third at Port C. At Port A, two passengers will be selected at random; each winner will receive one gift certificate worth $50. At Port B, one passenger will be selected at random to receive a gift certificate worth $35, and at Port C, one passenger will be selected at random to receive a gift certificate worth $25. How many different ways can the gift certificates be given out?

A. 270
B. 450
C. 2,700
D. 4,500
E. 27,000

Kudos for a correct solution.

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Re: Three small cruise ships, each carrying 10 passengers, will dock  [#permalink]

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New post 07 Jul 2015, 05:01
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Bunuel wrote:
Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at Port B, and the third at Port C. At Port A, two passengers will be selected at random; each winner will receive one gift certificate worth $50. At Port B, one passenger will be selected at random to receive a gift certificate worth $35, and at Port C, one passenger will be selected at random to receive a gift certificate worth $25. How many different ways can the gift certificates be given out?

A. 270
B. 450
C. 2,700
D. 4,500
E. 27,000

Kudos for a correct solution.


Number of ways 3 ships can go to 3 ports = 3! =6

Let ship A goes to dock A . Thus, the number of ways of selecting 2 winners out of 10 passengers = 10C2 = 45

Let ship B goes to dock B . Thus, the number of ways of selecting 1 winner out of 10 passengers = 10C1 = 10

Let ship C goes to dock C . Thus, the number of ways of selecting 1 winner out of 10 passengers = 10C1 = 10

Thus the total number of arrangements possible = 6*45*10*10 = 27000, thus E is the correct answer.
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Re: Three small cruise ships, each carrying 10 passengers, will dock  [#permalink]

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New post 07 Jul 2015, 05:59
Bunuel wrote:
Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at Port B, and the third at Port C. At Port A, two passengers will be selected at random; each winner will receive one gift certificate worth $50. At Port B, one passenger will be selected at random to receive a gift certificate worth $35, and at Port C, one passenger will be selected at random to receive a gift certificate worth $25. How many different ways can the gift certificates be given out?

A. 270
B. 450
C. 2,700
D. 4,500
E. 27,000

Kudos for a correct solution.


First Cruise Ship: 2 out of 10, order doesn't matter / no repetition > 10!/8!*2! = 45
Second Cruise Ship: 1 out of 10, order doesn't matter / no repetition > 10!/9! = 10
Second Cruise Ship: 1 out of 10, order doesn't matter / no repetition > 10!/9! = 10

Alltogether: 45*10*10 = 4500

Answer D
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Re: Three small cruise ships, each carrying 10 passengers, will dock  [#permalink]

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New post 07 Jul 2015, 09:32
1
Bunuel wrote:
Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at Port B, and the third at Port C. At Port A, two passengers will be selected at random; each winner will receive one gift certificate worth $50. At Port B, one passenger will be selected at random to receive a gift certificate worth $35, and at Port C, one passenger will be selected at random to receive a gift certificate worth $25. How many different ways can the gift certificates be given out?

A. 270
B. 450
C. 2,700
D. 4,500
E. 27,000

Kudos for a correct solution.


@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
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Re: Three small cruise ships, each carrying 10 passengers, will dock  [#permalink]

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New post 13 Jul 2015, 04:04
Bunuel wrote:
Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at Port B, and the third at Port C. At Port A, two passengers will be selected at random; each winner will receive one gift certificate worth $50. At Port B, one passenger will be selected at random to receive a gift certificate worth $35, and at Port C, one passenger will be selected at random to receive a gift certificate worth $25. How many different ways can the gift certificates be given out?

A. 270
B. 450
C. 2,700
D. 4,500
E. 27,000

Kudos for a correct solution.


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.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Three small cruise ships, each carrying 10 passengers, will dock  [#permalink]

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