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Permutation and Combinations

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Permutation and Combinations [#permalink] New post 05 Jan 2014, 18:00
I have a couple of questions on P n C and it will be great if someone can help me out with the concept/ process of solving.

In P n C, the questions involving pairs always knocks me out and I guess its because of lack of concept. So some additional problems/approach can help me tune up my skills in this area, and your help is highly appreciated.

Below are the questions:

1) The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match:
a) 210 b) 420 c) 840 d) None

Assumed correct answer is 420 and here is how it has been reached:

Choose two husbands out of 7: 7C2 = 21
Choose two wives <leaving aside the wives of the selected husbands>: 5C2 = 10
No. of teams = 21 x 10 = 210.
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 210 x 2 =420.

Pretty easy to understand and I have no problem accepting it. However, the problem arises when I go by pairs and here is my question to you why the below solution is not working out:
Team A:
Select a husband or wife from 7 pairs = 7 x 2 = 14
Select the other spouse <not the spouse of the selected husband> from 6 pairs = 6

Team B: Select a husband or wife from 5 pairs = 5 x 2 = 10
Select the other spouse <not the spouse of the selected husband> from 4 pairs = 4

Total selections = 14 x 6 x 4 x 10 = 84 x 40 = 3360
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 3360 x 2 = 6720
Can someone explain why the above approach is incorrect?
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Re: Permutation and Combinations [#permalink] New post 06 Jan 2014, 16:21
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AGH wrote:
I have a couple of questions on P n C and it will be great if someone can help me out with the concept/ process of solving.

In P n C, the questions involving pairs always knocks me out and I guess its because of lack of concept. So some additional problems/approach can help me tune up my skills in this area, and your help is highly appreciated.

Below are the questions:

1) The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match:
a) 210 b) 420 c) 840 d) None

Assumed correct answer is 420 and here is how it has been reached:

Choose two husbands out of 7: 7C2 = 21
Choose two wives <leaving aside the wives of the selected husbands>: 5C2 = 10
No. of teams = 21 x 10 = 210.
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 210 x 2 =420.

Pretty easy to understand and I have no problem accepting it. However, the problem arises when I go by pairs and here is my question to you why the below solution is not working out:
Team A:
Select a husband or wife from 7 pairs = 7 x 2 = 14
Select the other spouse <not the spouse of the selected husband> from 6 pairs = 6

Team B: Select a husband or wife from 5 pairs = 5 x 2 = 10
Select the other spouse <not the spouse of the selected husband> from 4 pairs = 4

Total selections = 14 x 6 x 4 x 10 = 84 x 40 = 3360
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 3360 x 2 = 6720
Can someone explain why the above approach is incorrect?

Dear AGH
I'm happy to respond. :-)

First of all, here's a blog post I think you will find helpful:
http://magoosh.com/gmat/2013/difficult- ... -problems/

One of the hardest things about counting question is to recognize when you are counting redundancies --- that is, when you are using a counting method that allows for the same combination to be counted in multiple ways. If your calculation produces a number much bigger than the OA, then it's always the case that you are are counting redundancies --- i.e., the same situation gets counted more than once.

Let's say the seven husbands are {a, b, c, d, e, f, g} and the seven corresponding wives are {A, B, C, D, E, F, G}.

Now, consider your method:
first choice = one of fourteen people
second choice = one of six non-spouses of first choice
third choice = one of the ten people in unused couples
fourth choice = one of four remaining non spouses

Consider the mixed doubles match that results from the four selections: {e, B, G, d}
Those same two teams would also be created by the selections:
{B, e, G, d}, {e, B, d, G}, {B, e, d, G}, {G, d, e, B}, {d, G, e, B}, {G, d, B, e}, and {d, G, B, e}.
Right away, that shows eight different ways that the same two teams can be selected using your approach, which means your answer of 3360 is 8 times too big. 3360/8 = 420, because it already would include all the swapped teams as separate choices.

Once again, counting problems are tricky: you have to be extremely careful that you are not counting the same thing more than once. The perspective you take in framing the counting problem is crucial.

Does all this make sense?
Mike :-)
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Mike McGarry
Magoosh Test Prep

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Re: Permutation and Combinations [#permalink] New post 06 Jan 2014, 19:29
Thanks a lot! your explanation makes it easier to understand. :-D
Re: Permutation and Combinations   [#permalink] 06 Jan 2014, 19:29
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