donut distribution

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Mon Nov 16, 2009 7:55 am

Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040

OA is A...why not C

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Source: Beat The GMAT — Problem Solving | Mon Nov 16, 2009 7:55 am

palvarez: Master | Next Rank: 500 Posts; Posts: 199; Joined: Sat Oct 24, 2009 4:43 pm; Thanked: 22 times; GMAT Score:710

by palvarez » Mon Nov 16, 2009 9:52 am

a +b +c = 5 0 <= a, b, c <=5

This equation has 7C2 solutions = 21.

This is a combination problem with repetition.

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Mon Nov 16, 2009 10:29 am

could you please elaborate on how you arrived at the 7C2 conclusion?

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palvarez: Master | Next Rank: 500 Posts; Posts: 199; Joined: Sat Oct 24, 2009 4:43 pm; Thanked: 22 times; GMAT Score:710

by palvarez » Mon Nov 16, 2009 10:39 am

you are given XXXXX; in how many ways, can you insert two |'s in it? 7C2.

XX|XX|X
X|XXX|X

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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

by vivekjaiswal » Mon Nov 16, 2009 10:47 am

palvarez wrote:you are given XXXXX; in how many ways, can you insert two |'s in it? 7C2.

XX|XX|X
X|XXX|X

Hi,

May be I am really dumb or something

I am still not able to connect the dots out here...how does this explain the distribution of donuts in my question?

I had thought it would be like, one of the guy could get anything from 0 to 5 donuts. thus the first one can get a donut in 6 ways then the next in 5 and the next in 4 and thus a total of 120. Now i know this is wrong, but I would really appreciate if you could come up with a better explanation please

Cheers,
Vivek

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palvarez: Master | Next Rank: 500 Posts; Posts: 199; Joined: Sat Oct 24, 2009 4:43 pm; Thanked: 22 times; GMAT Score:710

by palvarez » Mon Nov 16, 2009 11:03 am

Treat X as donuts, the bar | as splitting.

XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last

Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.

Why 2 |s? For three people in the above example, you need 2 |'s

For n people, you need n -1.

x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n

Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Nov 16, 2009 11:39 am

vivekjaiswal wrote:
palvarez wrote:you are given XXXXX; in how many ways, can you insert two |'s in it? 7C2.

XX|XX|X
X|XXX|X
Hi,

May be I am really dumb or something
I am still not able to connect the dots out here...how does this explain the distribution of donuts in my question?

I had thought it would be like, one of the guy could get anything from 0 to 5 donuts. thus the first one can get a donut in 6 ways then the next in 5 and the next in 4 and thus a total of 120. Now i know this is wrong, but I would really appreciate if you could come up with a better explanation please

Cheers,
Vivek

The problem with your solution is that the events are dependent, not independent.

For example, if person 1 gets 5 donuts, then person 2 and person 3 automatically get 0 each.

Your solution would work if there were 6 unique items and we were distributing those items among 3 people, 1 each. In that case, there would be 6 objects for the first person, then 5 for the second, then 4 for the third, for 6*5*4 possible distributions.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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tgf: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Wed Jul 15, 2009 12:10 pm

by tgf » Fri Nov 20, 2009 11:07 am

palvarez wrote:Treat X as donuts, the bar | as splitting.

XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last

Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.

Why 2 |s? For three people in the above example, you need 2 |'s

For n people, you need n -1.

x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n

Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.

Could someone just clarify what the difference between this problem at the one that Stuart Kovinsky answer to on the laste page of this thread: ?

https://www.beatthegmat.com/combination-t41362.html

How is the "zeros" taken care of here? If ||XXXXX leaves all the donuts for the last person, then how is the two slots (||) in the same place counted in the combinations??

Thanks

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palvarez: Master | Next Rank: 500 Posts; Posts: 199; Joined: Sat Oct 24, 2009 4:43 pm; Thanked: 22 times; GMAT Score:710

by palvarez » Fri Nov 20, 2009 12:09 pm

tgf wrote:
palvarez wrote:Treat X as donuts, the bar | as splitting.

XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last

Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.

Why 2 |s? For three people in the above example, you need 2 |'s

For n people, you need n -1.

x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n

Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.
Could someone just clarify what the difference between this problem at the one that Stuart Kovinsky answer to on the laste page of this thread: ?

https://www.beatthegmat.com/combination-t41362.html

How is the "zeros" taken care of here? If ||XXXXX leaves all the donuts for the last person, then how is the two slots (||) in the same place counted in the combinations??

Thanks

Here is how you solve it: If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?

x+y+z = 9 1 <= x, y, z <=9

a +b +c = 6 0<= a, b, c <=6

#solutions to the latter equation gives the answer: 8C2 = 28

The method works when each variable is between 0 and k, where k = x+y+z. So, you need to change the equation a bit to accomodate that constraint.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Fri Nov 20, 2009 4:37 pm

tgf wrote:
palvarez wrote:Treat X as donuts, the bar | as splitting.

XX|X|XX, 2 donuts for the first person; 1 for the next; 2 for the last

Now the problem is: in how many ways can you shuffle 2 '|' in between 5 donuts.

Why 2 |s? For three people in the above example, you need 2 |'s

For n people, you need n -1.

x_1 + x_2 + ... +x_k =n
where 0 <= x_i < =n

Number of solutions: (n+k-1)C(k-1) = (n+k-1)Cn. Here you see k-1 +'s.
Could someone just clarify what the difference between this problem at the one that Stuart Kovinsky answer to on the laste page of this thread: ?

https://www.beatthegmat.com/combination-t41362.html

How is the "zeros" taken care of here? If ||XXXXX leaves all the donuts for the last person, then how is the two slots (||) in the same place counted in the combinations??

Thanks

That solution is possibly the quickest way to solve this particular problem:

we have two partitions and 5 Xs to arrange, so the total number of arrangements is:

7!/5!2! = 7*6/2 = 21

(We're using the permutations formula for duplicate items: n!/r!s!t!... in which n is the total number of items and r, s, t, ... stand for the number of duplicates; in this question we have 7 total, a 5 duplicate and a 2 duplicate.)

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tgf: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Wed Jul 15, 2009 12:10 pm

by tgf » Sat Nov 21, 2009 5:36 am

Ah, I think a get it:

Since zero donuts for 3-1=2 people is possible, two slots can be placed in the same space, yielding 5+2=7 total slots where 2 and 5 objects are identical, leading to total permutations of (or combinatons of donuts)

7!/2!5!=21

But this is exactly the same as the number of combinations you can form by placing 2 slots on 5 donuts + 2 possble zeros = 7 objects, which is

7C2=21

I didn't see how those to approaches where equal, but I think I got it now.

Thanks

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tgf: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Wed Jul 15, 2009 12:10 pm

by tgf » Sat Nov 21, 2009 6:22 am

Just one last question:

Whether the 5 donuts are identical or different is completely irrelevant when is it comes to the answer?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Sat Nov 21, 2009 7:10 am

tgf wrote:Just one last question:

Whether the 5 donuts are identical or different is completely irrelevant when is it comes to the answer?

Right, because we don't care who gets which donuts, just who gets how many donuts.

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Pooja Bhula: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Tue Nov 24, 2009 12:48 am

by Pooja Bhula » Wed Nov 25, 2009 3:43 am

I went through the explanation n still dont get it... i thought of using another method, where each donut has 3 options of going to Larry, Micheal or Doug, thus making it 3^5...but this gets me an answer of 243 that does not match...please help me why understand why this is wrong, and what to do...

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Sun Jul 03, 2011 12:14 am

Stuart -

I still dont understand - please pardon my limited knowledge but can you please explain step by step

Stuart Kovinsky wrote:
tgf wrote:Just one last question:

Whether the 5 donuts are identical or different is completely irrelevant when is it comes to the answer?
Right, because we don't care who gets which donuts, just who gets how many donuts.

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