Problem Solving

Hi all,
This problem is an example from MGMAT's Word Translation's book, Chapter 12 (Comb/Prob/Stat: Advanced Strategy), p. 188:

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?

I understand that this is a mashup of both permutation and combinations. Part of the question deals with a specific order of ships and ports -- permutation. The other part is dealing with the prizes handed out -- combination.

My question is, is this a wrong approach to solve the problem?
1. Port A has 2 decisions, Port B has 1 decision, and port C has 1 decision.
2. Using the "slot method": __ __ __ __
3. 10 people per ship, so Port A will have 10 and 9 .
Port B will have 10 , and Port C will have 10 .
4. Multiply: 10 * 9 * 10 * 10 = 9000.
5. Since there are 3 ports, multiply this by three: 9000 * 3 = 27,000.

This is the same answer in the book, but the way numbers are generated is complex, due to the formulas. Is this approach/logic acceptable?

Thanks!
--Rishi

There are three ships. each carrying 10 passengers. So the first ship could go to any of the ports A, B or C: 3 ways. Second will then have 2 choices. Last ship can go to only one port then: 1 way.

Thus, total number of ways of arranging 3 ships to 3 ports= 3*2*1.= 6
(In other words, no of ways of arranging 3 ships to 3 ports are 3P3)

Next, port A has to select 2 people from 10 to distribute the prizes: 10C2=45
Similarly Port B and port C have to select one each from 10-10 guys : 10C1 and 10C1.= 10 and 10

Thus total number of ways:
6*45*10*10 =27000.

Thanks Prashant for the explanation, but I am more interested in some comments about the slot method for this particular problem.

I remember Ron Purewal saying in one of the Thursday study halls that all of the counting problems on the test can be completed using the slot method, but not all of the problems can be easily completed using the nCr and nPr formulas.

Do you have any insights regarding how to solve using the slot method?

Thanks,
--Rishi

Deleted

The slot method is not really recommended for this problem but anyway.

For the ships to land at the three ports. We consider the 3 ports as three slots as follows
__ __ __

In port A any of the three ships can dock. So,3
In port B any of the remaining two can dock so 2
In port C only the last remaining ship will dock so 1

3*2*1=6

At port A two will be selected out of 10. We consider two slots for the winners as follows
__ __
In slot 1, any of the 10 can be selected so 10
In slot 2, any of the remaining 9 can be selected so 9. Also, since we've used slots we've considered the different sequences in which the winners will be picked i.e. we have used permutations instead of combinations. Order in which winners are picked is obviously of no importance. So to mitigate that we divide by 2! i.e. 10*9/2!=90/2=45

At port B only 1 out of 10 is going to be picked so 10 possible ways
At port C only 1 out of 10 is going to be picked so 10 possible ways

Multiplying all the possible ways together we have
6*45*10*10=27000

Hey !!

What is the difference anyway between the two solutions ?? At least in this case, both have the same meaning, don't they? You may give it a fancy name like "SLOT METHOD" if you want to..

Well, this is mathematics. There is no master formula that is applicable to all or any set.. There are types and sub types.. To say that one "SLOT METHOD" should be good for all the problems in combinations, at least i won't support this approach. Knowing different ways is always better for understanding. Who knows ? may be there was another better or simpler way to solve the problem than you did using the SLOT METHOD !!

Cheers !

____________________________________

If you find my posts helpful, do I not deserve to be said " Thank you" ?

Thanks Knight247!

I see what I did wrong when I was using the slot method (even though the answer came out to be the same).
What I did was not count the total port combinations (3*2*1) -- I just used 3, and I did not divide by the number of same combinations (10 * 9 instead of [10*9]/ 2!).

Prashant,
I totally agree with your statement that there is no master formula to all sets. The video I was quoting from is this: https://vimeo.com/11327864.
The reason Ron said that the slot method is preferable is because it is easy to use and it is flexible. At around 48:20 he explains that the formulas work really well in problems where there are no restrictions.
Definitely knowing different approaches is better for understanding! Then you can use whichever is quickest!

Thanks for the help!
--Rishi

Yes Guys,
I agree that there isn't a fastest way to arrive at a solution. But it is best to understand every possible method when solving Combinatorics and Probability problems to strengthen one's basics. Atleast that is my approach. And once i'm comfortable with all the possible approaches then I'll try and strenghten and become an expert in one particular method. I'm currently going thru the Veritas Combinatorics and Probability Guide, and even they recommend the same thing. Infact, they recommend writing down all the possibilities when the number of possibilities are few or manageable. This approach gives a firm foundation in this absolutely unconventional topic in Math. lol

Haha!
I admit to using the "write down all the possibilities when the number of possibilities are few or manageable" approach once in a while. Just a few days ago on the Magoosh practice problems there was this one:
"Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?"

I was able to count them all in under a minute (whereas the "Average Pace" was about 2 mins 19 seconds)!
Heheheh