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A professor will assign seven projects to three students. If
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Updated on: 05 Oct 2011, 05:28
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A professor will assign seven projects to three students. If two students each got 2 projects, and the other one got 3 projects, how many ways are possible? A professor assigns three projects to seven students, therefore, the students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible?
Originally posted by zura on 01 Oct 2011, 00:25.
Last edited by zura on 05 Oct 2011, 05:28, edited 1 time in total.



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Re: combinations
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01 Oct 2011, 07:44
for the second question: 7c2 *5c2*3c3 >210



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Re: combinations
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01 Oct 2011, 07:55
@zura
I'm not seeing how you got different answers for these problems... when I set them up they look the same to me, just with different names for the groupings.
Problem 1: Groups are called "Students" Items are called "Projects"
In how many ways can seven Items be distributed across three Groups if Group 1 contains three Items, and Groups 2 and 3 contains two Items each?
Problem 2: Groups are called "Projects" Items are called "Students"
In how many ways can seven Items be distributed across three Groups if Group 1 contains three Items, and Groups 2 and 3 contains two Items each?



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Re: combinations
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01 Oct 2011, 11:36
I solved the first problem like this:
7C2 * 5C2 * 3C3 21 * 10 * 1 210
Since there are 3 ways any one student can get 3 projects while the other 2 get 2 each (3C1), multiply the previous answer by 3 for the final answer.
210 * 3 = 630
This is different from the given answer though, so I may be missing something.



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Re: combinations
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01 Oct 2011, 12:34
I see... I hadn't considered accounting for the variability of which student would get additional projects.
It's apparent, however, that the questions as posed are not in their original wording. I wonder if the original wording is clearer regarding this variability?
Edit:
Also, you could make the same argument for question two: that the variability of which project gets additional students would have to be accounted for.



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Re: combinations
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04 Oct 2011, 13:13
I think the 610 should be 210, the questions are indeed not only 'similar' but structurally identical.



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Re: combinations
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04 Oct 2011, 20:07
Both the problems should have the same answer. I get 630 for both the problems. There are 3 ways to choose which group should have 3 projects assigned. There are 7c2 * 5c2 *3c3 ways to assign them.
So 3 * 7c2 * 5c2 *3c3 = 630.



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Re: combinations
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06 Oct 2011, 03:15
zura wrote: posted two similar questions in one thread . A professor will assign seven projects to three students. If two students each got 2 projects, and the other one got 3 projects, how many ways are possible? A professor assigns three projects to seven students, therefore, the students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible? The projects are all distinct and so are the students. So in both the cases, you are doing essentially the same. In the first case, you are assigning 7 things to 3 people and in the second case, you are assigning 7 people to 3 things. So in both the cases, you choose the person/thing that will get 3 things/person in 3C1 = 3 ways. Next, you choose 3 things/people out of 7 in 7C3 ways. Next, you choose 2 things/people out of the remaining 4 in 4C2 ways. 3C1 * 7C3 * 4C2 = 3*(7*6*5)/(3*2) * 6 = 630 ways
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Re: A professor will assign seven projects to three students. If
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16 Jan 2015, 03:40
zura wrote: A professor will assign seven projects to three students. If two students each got 2 projects, and the other one got 3 projects, how many ways are possible? A professor assigns three projects to seven students, therefore, the students will be divided two three groups, which has three, two, two students, respectively. How many ways are possible? Responding to a pm: Quote: I believe the solution to the question should be 7C3 x 3C1 x 4C2 x 2C1.
However you have answered the question as 7C3 x 3C1 x 4C2. The reason I believe we should multiply it by 2C1 is because we need to select the student who gets the 2 projects selected by 4C2.
Your immediate & elaborate response would be most appreciated.
This is the reason your logic doesn't work: The problem is with distributing 4 projects between 2 students such that each student gets 2 projects each. Say students are Sa and Sb Say projects are P1, P2, P3 and P4. 4C2 is the way you select 2 projects for Sa. Whatever is left is for Sb. Say, you select P1 and P2 for Sa. You have P3 and P4 for Sb. P1 and P3 for Sa. You have P2 and P4 for Sb. P1 and P4 for Sa. You have P2 and P3 for Sb. P2 and P3 for Sa. You have P1 and P4 for Sb. P2 and P4 for Sa. You have P1 and P3 for Sb. P3 and P4 for Sa. You have P1 and P2 for Sb. There are 6 ways which is 4C2. Look at what happens when you do 4C2*2C1. You select P1 and P2 for Sa. You have P3 and P4 for Sb. Then since you multiply by 2, you select P1 and P2 for Sb. You have P3 and P4 for Sa.In another case, you select P3 and P4 for Sa. You have P1 and P2 for Sb. Then since you multiply by 2, you select P3 and P4 for Sb. You have P1 and P2 for Sa.Note that you have double counted both cases. When you select 2 out of 4, you are selecting them for one particular person. So you do not multiply by 2C1.
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Re: A professor will assign seven projects to three students. If
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Re: A professor will assign seven projects to three students. If
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