gurpreetsingh.1982 wrote:Hi all,

Was working on the GMATPrep1 earlier and found this question (refer to attached picture).

Could you please share how you would approach this problem? I worked it out as follows:

Since you have 8 options available for the 1st committee member, and 6 for the 2nd, and 4 for the third, the total number of committees would be 8 x 6 x 4.
However, this is not even a selection here - perhaps I've used the wrong approach here. Appreciate anyone's help in this, thanks!

Say the couples are A1A2, B1B2, C1C2 & D1D2.

It is for sure that one member from any 3 of the 4 couple group is to be chosen.

No. of ways of choosing any 3 couple group is 4C3 = 4.

Choosing one member each from the 3 chosen group will be 2.2.2 = 8.

Total no. of ways committee can be formed = 4.8 = 32.

Shalabh's Quants wrote:

gurpreetsingh.1982 wrote:Hi all,

Was working on the GMATPrep1 earlier and found this question (refer to attached picture).

Could you please share how you would approach this problem? I worked it out as follows:

Since you have 8 options available for the 1st committee member, and 6 for the 2nd, and 4 for the third, the total number of committees would be 8 x 6 x 4.
However, this is not even a selection here - perhaps I've used the wrong approach here. Appreciate anyone's help in this, thanks!

Say the couples are A1A2, B1B2, C1C2 & D1D2.

It is for sure that one member from any 3 of the 4 couple group is to be chosen.

No. of ways of choosing any 3 couple group is 4C3 = 4.

Choosing one member each from the 3 chosen group will be 2.2.2 = 8.

Total no. of ways committee can be formed = 4.8 = 32.

Hi Shalabh,

I understand how you choose a 3 couple group, and using 4C3 = 4 that is 4 ways a 3 couple group can be chosen.
Now what I still don't understand is the 2nd step, choosing one member each from the 3 couple group, which you did 2x2x2. Could you explain this portion a bit?
Much appreciated, thanks.

Gurpreet,

The best way to tackle combination questions is first..forget the formula!

Okay, here's my way and I really think its fast and makes sense.

You were absolutely correct with your approach in that you have 8 for the first 6 for the second and 4 for the third because you can't repeat people who are married.

So here is where the "slot method" comes into play:

First determine the amount of people you need to select from a certain group-->that number is three as the problem states. Write out the slots on your piece of paper like this:

------ ------ ------

In the first slot, how many people can we select? This is where your thought process was correct, so we fill in the slots like this:

8 6 4
- - -

"8" is the first slot, "6" is the second slot, "4" is the third slot

Using the fundamental counting principle you can multiply these numbers together (8*6*4) to get 192; however, since we are "selecting only 3 from the group" we must divide by the "number of slots!"

There are 3 slots, so 3! equals 6..when you divide 192/6 you see the answer becomes 32. Alternatively when you pick this method up, you'll notice that the 6 on top and the 6 on the bottom will cancel out leaving you with 8*4 or 32.

Hope this helps!

Thrills4ever wrote:Gurpreet,

The best way to tackle combination questions is first..forget the formula!

Okay, here's my way and I really think its fast and makes sense.

You were absolutely correct with your approach in that you have 8 for the first 6 for the second and 4 for the third because you can't repeat people who are married.

So here is where the "slot method" comes into play:

First determine the amount of people you need to select from a certain group-->that number is three as the problem states. Write out the slots on your piece of paper like this:

------ ------ ------

In the first slot, how many people can we select? This is where your thought process was correct, so we fill in the slots like this:

8 6 4
- - -

"8" is the first slot, "6" is the second slot, "4" is the third slot

Using the fundamental counting principle you can multiply these numbers together (8*6*4) to get 192; however, since we are "selecting only 3 from the group" we must divide by the "number of slots!"

There are 3 slots, so 3! equals 6..when you divide 192/6 you see the answer becomes 32. Alternatively when you pick this method up, you'll notice that the 6 on top and the 6 on the bottom will cancel out leaving you with 8*4 or 32.

Hope this helps!

Hi Thrills4ever,

Would like to secure my understanding of the slots method - indeed, we are choosing only 3 for the committee, however, why then do you divide by "3!"?
Is this because there are 6 ways in which these 3 people can be selected (i.e. 8/6/4, 8/4/6, 6/4/8, 6/8/4, 4/8/6, 4/6/8)?
Correct me if I'm getting your approach correctly here - still a little bit confused, I'm afraid.

Thanks,
Gurpreet

Gurpreet,

you are indeed correct..by dividing by the number of "slots!" you eliminate double counting. This is not a permutation question, because the order in selecting married couples does not matter. Does that make sense? You see that 8/6/4 or 4/6/8 are actually the same.

Think of it this way: there are 4 married couples so 8 people total. Does it matter if you first select married couple A or married couple B or married couple C? No it does not.

Here is another question: Jim needs to select a committee of 3 people from a group of 8 people. How many different ways can he accomplish this? Using the slot method see if you can figure it out!

Hi Thrills4ever,

Thanks for clarifying. Much clearer now, and since order doesn't matter, we divide by "slots!".
About the other question:- using the slots method, it would mean (8*7*6)/3! = 56.
Same result can be seen when we work out the formula 8C3. Am I right this time?

Anurag, your method too is well understood. Appreciate it.

Cheers,
Gurpreet

Gurpreet,

You are correct

Thanks Thrills4ever

Its very effective way of thinking and getting answer quickly.