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# A committee of three students has to be formed. There are

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Re: A committee of three students has to be formed. There are [#permalink]
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scheol79 wrote:
A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

3
4
5
6
8

If Paul is in the committee : We must choose 2 out of 3 others (No Stuart) --> 3 ways
If Stuart is in the committee : No Paul & Hence no Jane, so the other 2 must be chose --> 1 way
If neither Paul nor Stuart are in the committee, 3 students left, but since no Paul, no Jane : No commitee possible

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Re: A committee of three students has to be formed. There are [#permalink]

case 1 :
stuart is in the commitee- means only joan and jessica qualify as per conditions
so 1 way

case 2.
PAUL is in the commitee:
Jane, Joan and Jessica are other options

hence choosing 3 out of 4 = 4C3 = 4

hence answer = 1+4 = 5
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Re: A committee of three students has to be formed. There are [#permalink]
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bblast wrote:

case 1 :
stuart is in the commitee- means only joan and jessica qualify as per conditions
so 1 way

case 2.
PAUL is in the commitee:
Jane, Joan and Jessica are other options

hence choosing 3 out of 4 = 4C3 = 4

hence answer = 1+4 = 5

As Spidy001 pointed out, your logic is absolutely fine. You messed up in the last step. It should be 'hence choosing 2 out of 3 (out of Jane, Joan and Jessica)'. We have already chosen Paul.
3C2 = 3

hence answer = 1+3 = 4
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Re: A committee of three students has to be formed. There are [#permalink]
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I'm almost 16 years out of school, I found this one stumped me for a bit.

My preferred approach is to take all possibilities into account, and then backing out the restrictions.

Step (1)

We must choose 3 people of 5, easy enough 5c3 = 10 total possibilities

Step (2)

Looking at the first restriction, Paul and Stuart do not want to be on the committee together. So let's assume the breaking scenario where Paul or Stuart is selected. That gives us 3c2 scenarios, since we've already selected Paul or Stuart and the other is off the list.

So 3c2 = 3 possibilities that should be eliminated

Step (3)

Jane will only be on the committee with Paul. So we take all the scenarios will Jane will refuse. So if Jane is on the committee, and Paul may not be selected we have

3c2 = 3 possibilities should be eliminated

Summary

10 original possibilities
- 3 Paul/Stuart restriction
- 3 Jane/Paul restriction
===
4 remaining possibilities = the answer
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Re: A committee of three students has to be formed. There are [#permalink]
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senior wrote:
I'm almost 16 years out of school, I found this one stumped me for a bit.
My preferred approach is to take all possibilities into account, and then backing out the restrictions.

Let's try a different problem with your approach. There are five people: A, B, C, D, E. Need to chose 3 for a committee. A and B cannot be chosen together. B and C cannot be chosen together. How many options?

Your approach: total 10 options, 5c3.
Now, assume the wrong scenario where A and B are chosen together. There are three such scenarios. (A and B are chosen, just need one more person.) So we have to subtract the three wrong options. Similarly, there are three wrong scenarios where B and C are chosen together.

This gives us 10-3-3=4 as the answer.
Yet this answer is wrong. There are five possibilities: ACD, ACE, ADE, BDE, CDE.
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Re: A committee of three students has to be formed. There are [#permalink]
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SergeyOrshanskiy wrote:
Let's try a different problem with your approach. There are five people: A, B, C, D, E. Need to chose 3 for a committee. A and B cannot be chosen together. B and C cannot be chosen together. How many options?

Your approach: total 10 options, 5c3.
Now, assume the wrong scenario where A and B are chosen together. There are three such scenarios. (A and B are chosen, just need one more person.) So we have to subtract the three wrong options. Similarly, there are three wrong scenarios where B and C are chosen together.

This gives us 10-3-3=4 as the answer.
Yet this answer is wrong. There are five possibilities: ACD, ACE, ADE, BDE, CDE.

I think you've changed the problem slightly. What you've introduced with your new set of restrictions, is a situation where the restrictions are not mutually exclusive. That is to say, you're counting one of your exclusions twice and it would need to be added back in. When you do 5C3 for both sets, you get ABC as a restriction both times. You simply need to add that duplicate back in as your final step.

Originally posted by senior on 29 Jan 2013, 20:25.
Last edited by senior on 30 Jan 2013, 04:32, edited 1 time in total.
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Re: A committee of three students has to be formed. There are [#permalink]
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senior wrote:

I think you've changed the problem slightly. What you've introduced with your new set of restrictions, is a situation where the restrictions are not mutually exclusive. That is to say, you're counting one of your exclusions twice and it would need to be added back in. When you do 5C3 for both sets, you get ABC as a restriction both times. You simply need to add that duplicate back in as your final step.

Yes, you are right in your analysis. I think, the point Sergey was making was that this approach could make you falter if you are not extremely careful (he's totally right!). If there is an overlap in the two sets you are deducting out of the total, you need to take care to add that back.

Personally, I like the 'With B/without B' approach for such questions. There is no overlap possible since they are complementary situations.
With B - You cannot choose A and C so you must pick D and E. Only one way
Without B - You have 4 options and 4C3 ways to select the committee i.e. 4 ways
Total 1 + 4 = 5 ways
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Re: A committee of three students has to be formed. There are [#permalink]
VeritasPrepKarishma wrote:
I think, the point Sergey was making was that this approach could make you falter if you are not extremely careful...

Yes, this is the point I was trying to make. Even worse, if you have three restrictions, you would end up using an inclusion-exclusion formula, such as the one you are using to solve problems about overlapping sets...
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Re: A committee of three students has to be formed. There are [#permalink]
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SergeyOrshanskiy wrote:
Yes, this is the point I was trying to make. Even worse, if you have three restrictions, you would end up using an inclusion-exclusion formula, such as the one you are using to solve problems about overlapping sets...

Yes, you are right. In fact, I have found that sometimes sets concepts are extremely helpful in solving probability questions. I discussed an interesting sets method to solve a probability question in one of my posts sometime back.

@senior: You may like it so here is the link:
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Re: A committee of three students has to be formed. There are [#permalink]
VeritasPrepKarishma wrote:

I see. This way we do not have to count the arrangements with exactly one couple sitting together.

This was too complicated for me to think about, so I came up with a different solution.
First arrange two couples. There are three equiprobable arrangements: ([]), ([)], ()[].
In the first one there is a couple sitting together, so the stranger can "spoil" this arrangement in only one of five ways.
In the second one no couple is already sitting together, no matter what the stranger does.
In the third one there will be a couple sitting together, no matter what.
Thus we have 1/3*1/5 + 1/3 + 0 = 6/15=2/5.
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Re: A committee of three students has to be formed. There are [#permalink]
I have question here ..

with paul.. i think we shud take 2c2.. because

becase jane and paul will always be in one team because jane refuse to b in comitee without paul..and no staurt can be with them both.. so we have only choice jessica and joan..

correct me if m wrong
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Re: A committee of three students has to be formed. There are [#permalink]
sanjoo wrote:
I have question here ..

with paul.. i think we shud take 2c2.. because

becase jane and paul will always be in one team because jane refuse to b in comitee without paul..and no staurt can be with them both.. so we have only choice jessica and joan..

correct me if m wrong

Jane refuses to be in the committee without Paul but Paul doesn't refuse to be in the committee without Jane. This means that When Paul is in the committee, Jane may or may not be there. She is one of the candidates we will consider but we needn't 'have to have her in the committee'.

When Paul is not in the committee then Jane is not to be considered.
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Re: A committee of three students has to be formed. There are [#permalink]
So just to get this right and move on.

Is it incorrect to do 10C3 - 3C2 (Two of them can't be together) - 3C2 (Jane can only be with Paul, so let's pick the cases in which she needs 1 more member, Paul excluded) = 4

I got to the right answer but not sure if the logic is sound because I've never dealt with two parallel restrictions

Cheers
J
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Re: A committee of three students has to be formed. There are [#permalink]
jlgdr wrote:
So just to get this right and move on.

Is it incorrect to do 10C3 - 3C2 (Two of them can't be together) - 3C2 (Jane can only be with Paul, so let's pick the cases in which she needs 1 more member, Paul excluded) = 4

I got to the right answer but not sure if the logic is sound because I've never dealt with two parallel restrictions

Cheers
J

What you need to do is this:
5C3 - 3C1 - 3C2 = 4

5C3 (choose any 3 of 5)
3C1 (Ways of forming a committee having both Paul and Stuart. You need to select only 1 more from the 3 leftover people)
3C2 (Ways of forming a committee with Jane but without Paul. Select 2 of the leftover 3 people)

Note that 3C1 has Paul definitely and 3C2 does not have Paul definitely. Hence there will be no overlap in these committees.
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Re: A committee of three students has to be formed. There are [#permalink]
I just get the feeling that all these combinations are really misleading and unecessary to solve this one. Correct answer is 5 not 4 as official answer.

1 possible way - when Saul & Paul are out you have Jessica, Jane and Joan
3 ways when Sean is added to committe and combines: 1. Sean, Jessica, Joan 2. sean, jessica, jane & 3. Sean, joan, jane
1 way when paul is in the comitte - sean joan jessica

Adding those up it's 5! Am i crazy or what?

Posted from GMAT ToolKit
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Re: A committee of three students has to be formed. There are [#permalink]
oanaandries wrote:
I just get the feeling that all these combinations are really misleading and unecessary to solve this one. Correct answer is 5 not 4 as official answer.

1 possible way - when Saul & Paul are out you have Jessica, Jane and Joan
3 ways when Sean is added to committe and combines: 1. Sean, Jessica, Joan 2. sean, jessica, jane & 3. Sean, joan, jane
1 way when paul is in the comitte - sean joan jessica

Adding those up it's 5! Am i crazy or what?

Posted from GMAT ToolKit

The committees in red are not possible: Jane cannot be in the committee without Paul.

There are only 4 committees possible:
{Jane, Paul, Joan}
{Jane, Paul, Jessica}
{Paul, Jane, Jessica}
{Stuart, Joan, Jessica}

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Re: A committee of three students has to be formed. There are [#permalink]
This is indeed a very good question
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