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A committee of three students has to be formed. There are [#permalink]
 17 Nov 2010, 00:21
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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? A. 3 B. 4 C. 5 D. 6 E. 8
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Last edited by Bunuel on 08 Jan 2013, 10:52, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A committee of three students [#permalink]
17 Nov 2010, 00:48
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 Answer : 4
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Re: A committee of three students [#permalink]
17 Nov 2010, 05:32
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 Or you can look at it this way: There are two ways to make the committee: Way1: With PaulThen to get other two members, you can choose out of 3 people (Stuart cannot be chosen) = 3C2 = 3 Way 2: Without PaulThen Jane will not be in the committee. So three members needed and only three people to choose from.. 1 way Total ways of forming the committee = 3 + 1 = 4
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Re: A committee of three students [#permalink]
12 Jun 2011, 02:44
karishma-please let me know where i made a mistake : 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 [#permalink]
12 Jun 2011, 09:20
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bblast wrote: karishma-please let me know where i made a mistake : 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 [strike]hence choosing 3 out of 4 = 4C3 = 4[/strike] [strike]hence answer = 1+4 = 5 [/strike]
Paul is in the committee which means Stuart cannot be chosen. So we are left with 3 - Jane,Joan & Jessica. As we have already filled 1 seat with Paul, there are 2 seats left. Thus we are left with choosing 2 out of 3 which is 3c2 = 3 (not 4c3).
Hope its clear now.
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Re: A committee of three students [#permalink]
12 Jun 2011, 09:29
restriction 1 : if Paul is chosen then Stuart should not be chosen
if Stuart is chosen then Paul should not be chosen.
restriction 2 : No Paul no Jane
combining these two we have
case 1 : if Paul is chosen then Stuart should not be chosen
1c1 * 3c2 = 3
case 2 : if Stuart is chosen then Paul should not be chosen. Jane should be also left out .
1c1 *2c2 = 1
total possible arrangements = 3 +1 = 4.
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Re: A committee of three students [#permalink]
12 Jun 2011, 15:06
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bblast wrote: karishma-please let me know where i made a mistake : 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 [#permalink]
13 Jun 2011, 00:25
jo,pa,jes; pa,ja,jes; pa,ja,jo; total 3 combinations with paul. st,jes,jo; one combination with stuart. total 4.
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Re: A committee of three students [#permalink]
13 Jun 2011, 03:52
+1 spidy and thanks karishma
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Re: A committee of three students [#permalink]
08 Jan 2013, 05:12
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.
So we start with all possibilities:
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 [#permalink]
29 Jan 2013, 12:22
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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. Do you see your mistake?
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Re: A committee of three students [#permalink]
29 Jan 2013, 20:25
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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.
Do you see your mistake? 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.
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 [#permalink]
29 Jan 2013, 20:51
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 [#permalink]
29 Jan 2013, 21:22
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 [#permalink]
29 Jan 2013, 21:36
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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: http://www.veritasprep.com/blog/2012/01 ... e-couples/
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Re: A committee of three students [#permalink]
29 Jan 2013, 22:28
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
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29 Jan 2013, 22:28
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