PS: permutation

[amitdgr]

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

[Stuart@KaplanGMAT]

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

Let's try this one without using a formula.

For our first person, we have 10 choices.

Once we've chosen a person, we also eliminate his or her spouse. So, for our second person, we have 8 choices.

Now we've eliminated both of those people and their spouses, so for our third person, we have 6 choices.

Therefore, we have 10*8*6 = 480 possible trios.

[Ian Stewart]

The phrasing of the question is not good- in mathematical counting, the word 'arrangements' implies that order matters, but there is no reason to think order should matter in this question. A real GMAT question will never be ambiguous about whether order matters.

If order matters, Stuart's solution above is correct. If order does not matter, you must divide 480 by 3! = 6, the number of ways to put the three people selected in different orders. So if order is not intended to matter, the answer should be 80.

Alternatively you could look at the problem as follows (assuming order does not matter):

-there are 10C3 = 120 ways to choose three people;
-if we choose a group of three which *does* include a married couple, we have five choices for the married couple, and 8 choices for the person to join them, or 5*8 = 40 choices in total;
-we thus must have 120-40 = 80 ways of choosing three people and *not* including a married couple.

[stop@800]

Great explanation IAn, tahnks a lot

I never used formulas in these question but now I think I will be able to...

MY method:
10 ppl

so selection of first from 10 ppl
so selection of second from 8 ppl
so selection of third from 6 ppl

Ways = 10 * 8 * 6 / 3!

[amitdgr]

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

Let's try this one without using a formula.

For our first person, we have 10 choices.

Once we've chosen a person, we also eliminate his or her spouse. So, for our second person, we have 8 choices.

Now we've eliminated both of those people and their spouses, so for our third person, we have 6 choices.

Therefore, we have 10*8*6 = 480 possible trios.

Thanks for the reply Stuart.. your method is very simple. I love your formula-less approach.

The phrasing of the question is not good- in mathematical counting, the word 'arrangements' implies that order matters, but there is no reason to think order should matter in this question. A real GMAT question will never be ambiguous about whether order matters.

I guess I messed up the question by mentioning "arrangements". I think it should be "selections".

Ian, how will a GMAT question be unambiguous about whether order matters ? Do they mention it explicitly ? Or are there any indicators to look out for ?

Alternatively you could look at the problem as follows (assuming order does not matter):

-there are 10C3 = 120 ways to choose three people;
-if we choose a group of three which *does* include a married couple, we have five choices for the married couple, and 8 choices for the person to join them, or 5*8 = 40 choices in total;
-we thus must have 120-40 = 80 ways of choosing three people and *not* including a married couple.

Did I understand this right ?

I understand from your solution that there are 5 ways of selecting a married couple ( since 5 pairs) and remaining one place will be filled by one of the 8 ppl. so it is 5*8 = 40 choices.


Thanks. You people make this forum the great place it is