Number of ways of Project Allocation

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goelmohit2002: Legendary Member; Posts: 1799; Joined: Wed Dec 24, 2008 3:03 am; Thanked: 36 times; Followed by:2 members

Number of ways of Project Allocation

by goelmohit2002 » Thu Aug 27, 2009 12:54 am

An instructor will assign seven projects to three students. If two students each got two projects each, and the other one got 3 projects. In how many ways can the instructor allocate the seven projects to three students.

[spoiler]OA = 630. Can someone please tell how ?[/spoiler]

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Source: Beat The GMAT — Problem Solving | Thu Aug 27, 2009 12:54 am

tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Thu Aug 27, 2009 1:10 am

trick is when you select the 2 students who get 2 books from the three students, you are automatically selecting the 3rd student who gets the 3 books, because there are only 3 books left.
Answer is:

7C2 *5C2 * 3C2= 630

The powers of two are bloody impolite!!

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goelmohit2002: Legendary Member; Posts: 1799; Joined: Wed Dec 24, 2008 3:03 am; Thanked: 36 times; Followed by:2 members

by goelmohit2002 » Thu Aug 27, 2009 1:33 am

tohellandback wrote:trick is when you select the 2 students who get 2 books from the three students, you are automatically selecting the 3rd student who gets the 3 books, because there are only 3 books left.
Answer is:

7C2 *5C2 * 3C2= 630

sorry I could not get your reasoning....

Based on your reasoning I guess the correct answer should have been like:

7C2 * 5C2 * 3C3

Can you please tell why are you selecting 2 from 3 books left at the end ?

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Thu Aug 27, 2009 1:40 am

goelmohit2002 wrote:
tohellandback wrote:trick is when you select the 2 students who get 2 books from the three students, you are automatically selecting the 3rd student who gets the 3 books, because there are only 3 books left.
Answer is:

7C2 *5C2 * 3C2= 630
sorry I could not get your reasoning....

Based on your reasoning I guess the correct answer should have been like:

7C2 * 5C2 * 3C3

Can you please tell why are you selecting 2 from 3 books left at the end ?

its not the books I am selecting. its the 2 guys I am selecting out of three so that they can get 2 books each.
The third guy obviously gets 3 books

Ok lemm explain:

1st guy gets two books= 7C2 ways
2nd guy gets 2 books= 5C2 ways-(because 5 books left)

third guys gets the three books left, nothing to do here

now those first two guys can be selected in 3C2 ways.
we don't need to select the third guy. because when we select the two guys, we automatically select the third guy who gets the remaining 3 books..

Last edited by tohellandback on Thu Aug 27, 2009 3:50 am, edited 1 time in total.

The powers of two are bloody impolite!!

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enniguy: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sat Jul 19, 2008 6:41 am; Thanked: 8 times

IMO 630

by enniguy » Thu Aug 27, 2009 3:25 am

<edited> 630 it is. </edited>

There are three students out of which one gets 3 books. Total number of ways to decide who gets 3 books = 3.

Now, a student can be given 3 books out of 7 books in
7C3 = 35 ways.

The next student can be given 2 books out of remaining 4 books in
4C2 = 6 ways.

Last one will get 2 books in 2 remaining books in
1 ways.

So,
<edited>
3 ( 35 * 6 * 1) = 630 ways.
</edited>

Last edited by enniguy on Thu Aug 27, 2009 6:35 am, edited 2 times in total.

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csabiga: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Mon Aug 10, 2009 6:19 am

Re: IMO 126

by csabiga » Thu Aug 27, 2009 3:41 am

enniguy wrote: So,
3 ( 35 + 6 + 1) = 126 ways.

These are not different situations.

The first student can get the books 35 ways AND after he/she got the books the second one can get from the others in 6 ways. Third student cannot choose.
You have to multiply the values.

3*(35*6*1) = 630 ways

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[email protected]: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Thu Aug 27, 2009 4:44 am

the answer

by [email protected] » Thu Aug 27, 2009 5:04 am

There are two selections : First of selecting the students and then selection of books.

Now we are not sure which student gets 2 books. So Let us choose which student gets 2 books. First student can be chosen in 3C1 way. Second student who gets 2 books can be chosen in 2C1 way. (as only 2 students remain for selection)

So there are 3C1 * 2C1 ways to choose students.

Now there are 7C2 ways to select 2 books given to student 1. And 5C2 ways to select 2 books to be given to student 2. And 3C3 ways to select 3 books to be given to student 3.

So total ways = 3C1 * 2C1 * 7C2 * 5C2 * 3C3 == 1260...

Let me know what is correct answer.