A committee of three people is to be chosen from four

[ifydaniel100]

Is it possible to use the "Glue Method" (for the married couple chosen) as taught in the Manhattan GMAT Math strategy guide?

Total committees of 3 that can be chosen from 8. 8C3 = 56

Now for the committees that involve the married couples - which I don't want - I thought to use the 'glue method'. Instead of 3 slots available, I made only 2; one that holds the married couple, and the other that holds one half of any of the other couples.

*
Choose 2 from 4 couples. 4C2 = 12. Next multiply this by 2 because the couple in one of the slots could be arranged in reverse.
12 x 2 = 24.
*

Finally 56 - 24 = 32

I don't feel confident about the section in asterisks because
1. I could just as easily have said "choose 2 from 8 people" which would have been wrong
2. Rearranging the couple shouldn't matter if order doesn't matter

Lastly, if the glue method cannot be used here can you provide an example in which its use is most effective? Thank you.

[tim]

the reason you multiply 12 by 2 is not because there are two ways to order the married couple - remember you're just choosing a committee, so it doesn't matter who gets chosen first or second. the real reason you multiply by 2 is because the couple that is only represented by one member on the committee can be represented by either the husband or the wife*.

*(apologies if the heteronormativity offends anyone; these just seemed the easiest terms to use to get the point across)

[ifydaniel100]

I just realized I made an error in my calculation. The section in asterisks I wrote 4C2 is 12. It is actually 6. Now I'm not sure how to approach this with the glue method strategy. Please assist. Thank you!

[jnelson0612]

I just realized I made an error in my calculation. The section in asterisks I wrote 4C2 is 12. It is actually 6. Now I'm not sure how to approach this with the glue method strategy. Please assist. Thank you!

The glue method works! Let's examine how we can use it successfully here:

1) You have correctly determined that if we just choose 3 people out of 8, disregarding the constraint that we cannot have married couples, we get 56 possible committees. This number includes acceptable committees (no married couples) and unacceptable committees (those including a married couple).

2) Now, let's glue the married couples together. Let's call them AB, CD, EF, and GH. Let's use this method to figure out how many unacceptable committees could be create:
Pair AB can combine with C, D, E, F, G, or H. 6 possible committees.
Pair CD can combine with A, B, E, F, G, or H. 6 possible committees.
Pair EF can combine with A, B, C, D, G, and H. 6 possible committees.
Obviously Pair GH can create 6 possible committees also.

Thus, we have 24 unacceptable committees.

3) Let's subtract 24 unacceptable committees of the 56 total committees. That leaves us with 32 acceptable committees. :-)

I do believe that the slot method is a little quicker and easier but this method is certainly valid too.

[dina_84]

can i choose 1 person from each couple, so that none of the couple is in the same committee then choose 3 person from it?
2C1*2C1*2C1*2C1*4C3=64
but it does not give the same answer for when i choose 3 couple from 4 of the couple then choose 1 person from each couple
4C3*2C1*2C1*2C1=32
why?
thanks

[RonPurewal]

can i choose 1 person from each couple, so that none of the couple is in the same committee then choose 3 person from it?
2C1*2C1*2C1*2C1*4C3=64

If you do this, you're counting each possible composition of the committee twice.

E.g., consider the possibility in which the committee consists of wife #1, wife #2, and wife #3.
With this method, you're going to get this committee twice -- once when you choose all four wives as your "starting group", and again when your "starting group" is wife #1, wife #2, wife #3, and husband #4.
The same thing is going to happen every time, because the fourth person (the one NOT chosen from the starting group) could be either of two people.

but it does not give the same answer for when i choose 3 couple from 4 of the couple then choose 1 person from each couple
4C3*2C1*2C1*2C1=32
why?
thanks

This is a valid approach, because it will only give each possible composition once.

[jtslbr]

I was trying to solve this problem using the counting method.
there are 4 couples so total 8 people
We need a 3 member committee where none of the 3 are a couple

SO 1st person can be chosen in 8 ways
2nd person can be chosen in 8-2 = 6 ways
and the 3rd person can be chosen in 8-2-2 = 4 ways

But my method gets me 8*6*4 which is wrong..

What am I missing here?

yes, this is the simplest way.

you should be able to solve any combinatorics problem with the "slot" (multiplication) method.
you may have to divide by a factorial or factorials (if order doesn't matter), but, other than that, the slot method should pretty much always work.

in this problem:
as the poster has noted above, there are 8 choices for the first person, 6 for the second, and 4 for the third.
but ORDER DOESN'T MATTER, so you have to divide by the FACTORIAL OF THE NUMBER OF THINGS CHOSEN:
(8 x 6 x 4) / 3!
= 32

why do you divide by 3! and not 5!=(8-3)!

[tim]

The very last sentence of the post you quoted contains your answer!

[RonPurewal]

The very last sentence of the post you quoted contains your answer!

Yep.

Also notice that division by 5! gives a non-integer answer, so you can reject it out of hand. (The answer to "How many ways..." clearly must be a whole number.)