Committee question

[gurpreetsingh.1982]

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!

[Shalabh's Quants]

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.

[gurpreetsingh.1982]

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.

[Thrills4ever]

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!

[gurpreetsingh.1982]

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

[Thrills4ever]

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!

[Anurag@Gurome]

Number of ways to choose 3 people out of 4 = 4C3
Number of different committees that can be chosen if two married people, both cannot serve the committee = 4C3 * 2^3 = 4 * 8 = 32 (2^3, as we can choose any 2 people from each chosen team)

The correct answer is E.

[gurpreetsingh.1982]

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

[Thrills4ever]

Gurpreet,

You are correct

[krishna239455]

Thanks Thrills4ever

Its very effective way of thinking and getting answer quickly.