fangtray wrote:I think the problem I'm having understanding this area of the Quant portion is that it seems like everytime I've learned something, another problem changes it so its as if the rules change. Consider the following problem:

Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at port B, and another at Port C. At Port A, 2 passengers will be selected at random, each winner will receive one gift certificate worth $50. At port B, 1 passenger will be selected at random to receive a gift certificate worth $35. and at port C, one passenger will be selected to receive a gift certificate worth $25. How many different ways can the gift certificates be given out?


First we have to calculate how many different ways the ships can dock. And this is the part one of the parts I can't get past. 3 ships, 3 docks. Why isn't it 3 x 3?

The wording seems a little vague to be honest. When they ask for how many different ways to distribute the gift certificates, I'm assuming that it's based on the amounts and not the distinct certificates themselves.

If that's the case, then let's start by breaking up the people into three groups, with 10 people in each...

Now, there are 3! = 6 possible port configurations for the three ships. To help visualize this, we can treat each port as a character position within a three-letter word, with each ship being a different letter. In other words:

ABC
ACB
BAC
BCA
CAB
CBA

Now we need to calculate the number of ways to distribute the gift certificates for each port. For port A, we have 10C2 = 45 ways. For port B, there are 10C1 = 10 ways. For port C, there are also 10C1 = 10 ways.

Since each of these events is independent, we simply multiply them: 45*10*10 = 4500 ways

Finally, we multiply this result by 6: one for each possible ship-to-port configuration.

4500*6 = 27000 ways

Consider it this way:-

3 ships A B and C , 3 docks D1 D2 D3

Ship A can choose, dock in 3 ways, once chosen lets go to ship B, it can now choose in only two ways , as one dock is already occupied by A, thereafter C has only one choice left. Thus 3 X 2 X 1 = 6. Now the solution to the problem:-

6 ways X 10 C 2 X 10 C 1 X 10 C 1 total ways

6 ways for the ships to dock

X 10 C 2 for choosing 2 guys from ship A

X 10 C 1 Thereafter choosing 1 guy from ship B

X 10 C 1 choosing 1 guy from ship C

6 ways X 10 C 2 X 10 C 1 X 10 C 1 total ways = 6x10 x9x10x 10/2 = 27000

Hope thats thee right answer and also it helps!

fangtray wrote:I think the problem I'm having understanding this area of the Quant portion is that it seems like everytime I've learned something, another problem changes it so its as if the rules change. Consider the following problem:

Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at port B, and another at Port C. At Port A, 2 passengers will be selected at random, each winner will receive one gift certificate worth $50. At port B, 1 passenger will be selected at random to receive a gift certificate worth $35. and at port C, one passenger will be selected to receive a gift certificate worth $25. How many different ways an the gift certificates be given out?


First we have to calculate how many different ways the ships can dock. And this is the part one of the parts I can't get past. 3 ships, 3 docks. Why isn't it 3 x 3?

To answer your question about 3x3, I would suggest writing out the possibilities rather than trying to rely on formulas alone (especially when you're feeling uncertain). I can see where having 3 ports and 3 ships can be confusing, but in any arrangement problem (ie. how to arrange ships to ports), one of these numbers (ie. # of ports) will represent slots or spaces, while the other (ie. # of ships) will represent the objects to fill in those slots or spaces.

krusta80 wrote:

fangtray wrote:I think the problem I'm having understanding this area of the Quant portion is that it seems like everytime I've learned something, another problem changes it so its as if the rules change. Consider the following problem:

Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at port B, and another at Port C. At Port A, 2 passengers will be selected at random, each winner will receive one gift certificate worth $50. At port B, 1 passenger will be selected at random to receive a gift certificate worth $35. and at port C, one passenger will be selected to receive a gift certificate worth $25. How many different ways an the gift certificates be given out?


First we have to calculate how many different ways the ships can dock. And this is the part one of the parts I can't get past. 3 ships, 3 docks. Why isn't it 3 x 3?

To answer your question about 3x3, I would suggest writing out the possibilities rather than trying to rely on formulas alone (especially when you're feeling uncertain). I can see where having 3 ports and 3 ships can be confusing, but in any arrangement problem (ie. how to arrange ships to ports), one of these numbers (ie. # of ports) will represent slots or spaces, while the other (ie. # of ships) will represent the objects to fill in those slots or spaces.

thats what i did and my goodness i got 9.

SHIPS A, B and C. Docks 1, 2 and 3

A1, B2, C3,
A3, B1, C2
A2, B3, C1

what the?! Isn't it like..when there's 3 dinner entrees and 3 desserts, how many different combinations can you have? and its 3 *3?

fangtray wrote:I think the problem I'm having understanding this area of the Quant portion is that it seems like everytime I've learned something, another problem changes it so its as if the rules change. Consider the following problem:

Three small cruise ships, each carrying 10 passengers, will dock tomorrow. One ship will dock at Port A, another at port B, and another at Port C. At Port A, 2 passengers will be selected at random, each winner will receive one gift certificate worth $50. At port B, 1 passenger will be selected at random to receive a gift certificate worth $35. and at port C, one passenger will be selected to receive a gift certificate worth $25. How many different ways an the gift certificates be given out?


First we have to calculate how many different ways the ships can dock. And this is the part one of the parts I can't get past. 3 ships, 3 docks. Why isn't it 3 x 3?

Below is an alternate approach.

Number of options for the prize winner at Dock C = 30. (Any of the 30 passengers on the 3 ships.)
Number of options for the prize winner at Dock B = 20. (Any of the 20 passengers on the remaining 2 ships.)
Number of options for the pair of winners at Dock A = 10C2 = 45. (Any pair that can be formed from the 10 passengers on the one remaining ship.)
To combine these options, we multiply:
30*20*45 = 27,000.