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Difference between two combinatorics problems

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Difference between two combinatorics problems [#permalink] New post 30 Dec 2013, 14:15
Hi everyone,

I don't get the difference between the following two problems:

1) Six highschool boys gather at the gym for a modified game of basketball involving three teams.
Three teams of 2 people each will be created. How many ways are there to create these 3
teams?

(A)15
(B)30
(C)42
(D)90
(E)108

2) How many possible ways can 3 girls (Rebecca, Kate, Ashley) go on a date with 3 boys
(Peter, Kyle, Sam)?

(A)3
(B)4
(C)5
(D)6
(E)8

The corresponding ways to solve the problems are as follows:
1) 10!/(5!*2!) = 15
We divide by 2! as there are two teams - A and B - and it does not matter whether it is A and B or B and A.

2) 3*2*1 = 6
Originally I got it wrong and after seeing the correct answer the slot method came to my mind - For the first lady there are 3 men, for the second there are 2 man left, etc. Actually, I am not sure whether or not this is right. The count method looks more appropriate now that i am writing the post - 3 + 2 + 1 (using the same logic as above).

However, I don't seem to get why we do not divide by 3! as there is no difference in whether it is A, B and C or C, B and A, etc. (e.g. the order of the couples which is 3! as there are 3 couples). From what I've seen so far this way of solving a problem (by dividing by the number of groups) is applicable when using the slot and the combinatorics formula method.

Any help is truely appreciated.
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Re: Difference between two combinatorics problems [#permalink] New post 02 Jan 2014, 12:01
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mihailesko wrote:
Hi everyone,

I don't get the difference between the following two problems:

1) Six highschool boys gather at the gym for a modified game of basketball involving three teams.
Three teams of 2 people each will be created. How many ways are there to create these 3
teams?

(A)15
(B)30
(C)42
(D)90
(E)108

2) How many possible ways can 3 girls (Rebecca, Kate, Ashley) go on a date with 3 boys
(Peter, Kyle, Sam)?

(A)3
(B)4
(C)5
(D)6
(E)8

The corresponding ways to solve the problems are as follows:
1) 10!/(5!*2!) = 15
We divide by 2! as there are two teams - A and B - and it does not matter whether it is A and B or B and A.

2) 3*2*1 = 6
Originally I got it wrong and after seeing the correct answer the slot method came to my mind - For the first lady there are 3 men, for the second there are 2 man left, etc. Actually, I am not sure whether or not this is right. The count method looks more appropriate now that i am writing the post - 3 + 2 + 1 (using the same logic as above).

However, I don't seem to get why we do not divide by 3! as there is no difference in whether it is A, B and C or C, B and A, etc. (e.g. the order of the couples which is 3! as there are 3 couples). From what I've seen so far this way of solving a problem (by dividing by the number of groups) is applicable when using the slot and the combinatorics formula method.

Any help is truly appreciated.

Dear mihailesko,
That's a great question, and I am happy to help. :-)

First of all, the questions are very different, because in the first case, any two elements of the six can be paired on an individual team, whereas in the second problem that's not the case --- each pair must be a male and a female, not two males or two females.

For the first question, the expression you gave, 10!/(5!*2!), does not equal 15 --- as a matter of fact, it equals 15120, a number far bigger than anything appropriate for the problem. Instead, here's how I would think about that problem. Suppose the six people are A, B, C, D, E, F.

Step #1 --- consider all permutations of all six --- 6! --- one might be F, C, E, B, A, D, which would correspond to the teams (F, C), (E, B), and (A, D)
Now, we have to eliminate redundancies
Step #2 --- within each pair separately, we could switch the order (e.g. (C, F) instead of (F, C)), and the team would be the same. Thus, we have to divide by 2 for each time --- divide by 2^3 = 8
Step #3 --- The same teams could be rearranged in a different order: for example, (A, D), (F, C), and (E, B). For the same three teams, we could rearrange them in 3! = 6 orders, so we need to divide by this as well:
6!/(8*3!) = (6*5*4)/8 = 3*5 = 15

That's how I would do the first one.

For the second problem, what you ask is a very subtle question. You see, everything about counting problems is very subtle. It's never about just applying a formula or method --- if that is your approach, you will not master these problems. Counting problems are always about framing the problem correct --- that is, asking the right questions, and viewing the question the right way. In the solution to the first problem, I framed the problem in a particular way that made the calculation easy. That was not the only way to solve, but it seemed to me the most efficient solution. For the second problem, it simply doesn't make sense to consider a "first couple", and then "second couple", and then later, remove all the redundancies. It is considerably more efficient just to say the three females are "constants", and randomly distribute the three males to the three fixed females. Unlike the first problem, we know that three elements, Rebecca, Kate, and Ashley, will never be paired with each other and in fact can be used to define the three couples ---- we know, after the selection, there will be one couple that includes Rebecca, one that includes Kate, and one that includes Ashley, and in any selection result, those three always will define the complete set of couples. That's precisely why we take advantage of that when we design our solution by considering the three females as "fixed constants".

Here's a blog you may find helpful:
http://magoosh.com/gmat/2013/difficult- ... -problems/

Does all this make sense?
Mike :-)
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Re: Difference between two combinatorics problems [#permalink] New post 03 Jan 2014, 10:12
Mike,

Thanks for posting a reply. Now that I see it, I must have been out of space when I wrote the thread (I wrote the way to solve a different question. What's more, I also did that wrong! :D). Whatever.

Even though I feel ashamed you have managed to find the issue that made me write the post - seeing the subtle difference as you say - and explain it in a great way. I find this skill amazing.

I think I understood, but have a look on how my brain has processed the information:

Because the ladies are a constant, I somehow make a paralel to a problem where you have a four-digit number (let's say abcd), which has its first digit fixed and 3, 2, and 1 digits, respectfully, for the second, third and forth digit of the number.

Therfore, the # of the different four-digit numbers, which begin with A and have B, C and D as the other digits, is:

ABCD -> 1*3*2*1 = 6

Am I correct?

I highly recommend the blog that you linked. You have done some "genuis play" there.

Thanks alot!
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Re: Difference between two combinatorics problems [#permalink] New post 03 Jan 2014, 12:42
Expert's post
mihailesko wrote:
Mike,
I think I understood, but have a look on how my brain has processed the information:

Because the ladies are a constant, I somehow make a parallel to a problem where you have a four-digit number (let's say abcd), which has its first digit fixed and 3, 2, and 1 digits, respectfully, for the second, third and forth digit of the number.

Therefore, the # of the different four-digit numbers, which begin with A and have B, C and D as the other digits, is:

ABCD -> 1*3*2*1 = 6

Am I correct?

I highly recommend the blog that you linked. You have done some "genius play" there.

Thanks a lot!

Dear mihailesko,
If the first number, the thousand's place, is fixed (I'm going to say 6, just to make it a different digit from the others), and in the other three places, the digits 1 & 2 & 3 appear in any possible order, then you are perfectly correct: there are 3! = 6 different numbers. Here they are, in numerical order:
6123
6132
6213
6231
6312
6321
If there are only a small number of possibilities, it can be very helpful to list them just to reinforce your understanding. Listing possibilities is often not enough to solve a GMAT combinatorics problem, but starting a list can often be a good strategy to help you decide what solution path to take.
Does all this make sense?
Mike :-)
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Mike McGarry
Magoosh Test Prep

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Re: Difference between two combinatorics problems   [#permalink] 03 Jan 2014, 12:42
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