donut distribution

[This topic has 3 expert replies and 10 member replies]
Forum Index GMAT Math Problem Solving
Free $100 Amazon.com Gift Card - Buy a GMAT course using a Beat The GMAT discount code between Mar 8-22 and get a $100 Amazon.com Gift Card. Learn more!
Post New Topic   Post Reply

vivekjaiswal
Really wants to Beat The GMAT!

Default Avatar

Joined: 15 Sep 2009
Posts: 107

Thanks given: 14
Thanked 6 times in 6 posts
Location: Pune, India

Test Date: Still Planning
Target GMAT Score: 730+
GMAT Score: 630
Topic: donut distribution
PostMon Nov 16, 2009 8:55 am
Elapsed Time:
00:00
 Lap   Why a timer is critical to improving your score
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
Back to top
View user's profile Send private message
palvarez
Really wants to Beat The GMAT!

Default Avatar

Joined: 24 Oct 2009
Posts: 199

Thanks given: 3
Thanked 19 times in 17 posts

GMAT Score: 710
PostMon Nov 16, 2009 10:52 am
a +b +c = 5 0 <= a, b, c <=5


This equation has 7C2 solutions = 21.

This is a combination problem with repetition.
Back to top
View user's profile Send private message
vivekjaiswal
Really wants to Beat The GMAT!

Default Avatar

Joined: 15 Sep 2009
Posts: 107

Thanks given: 14
Thanked 6 times in 6 posts
Location: Pune, India

Test Date: Still Planning
Target GMAT Score: 730+
GMAT Score: 630
PostMon Nov 16, 2009 11:29 am
could you please elaborate on how you arrived at the 7C2 conclusion?
Back to top
View user's profile Send private message
palvarez
Really wants to Beat The GMAT!

Default Avatar

Joined: 24 Oct 2009
Posts: 199

Thanks given: 3
Thanked 19 times in 17 posts

GMAT Score: 710
PostMon Nov 16, 2009 11:39 am
you are given XXXXX; in how many ways, can you insert two |'s in it? 7C2.


XX|XX|X
X|XXX|X
Back to top
View user's profile Send private message
vivekjaiswal
Really wants to Beat The GMAT!

Default Avatar

Joined: 15 Sep 2009
Posts: 107

Thanks given: 14
Thanked 6 times in 6 posts
Location: Pune, India

Test Date: Still Planning
Target GMAT Score: 730+
GMAT Score: 630
PostMon Nov 16, 2009 11:47 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 Smile
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 Smile

Cheers,
Vivek
Back to top
View user's profile Send private message
palvarez
Really wants to Beat The GMAT!

Default Avatar

Joined: 24 Oct 2009
Posts: 199

Thanks given: 3
Thanked 19 times in 17 posts

GMAT Score: 710
PostMon Nov 16, 2009 12:03 pm
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.
Back to top
View user's profile Send private message
Reply from GMAT/MBA Admissions Expert
See All Featured Experts
Stuart Kovinsky
GMAT Instructor

Most Active Expert Most Thanked Expert
Joined: 08 Jan 2008
Posts: 2263

Thanks given: 0
Thanked 541 times in 467 posts
Location: Toronto

GMAT Score: 800
PostMon Nov 16, 2009 12:39 pm
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 Smile
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 Smile

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, B.A. LL.B.
Academic Manager
Admissions Consultant
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST

GMAT Blogs

Learn more about me
Back to top
View user's profile Send private message
Thanked by: vivekjaiswal
tgf
Just gettin' started!

Default Avatar

Joined: 15 Jul 2009
Posts: 16

Thanks given: 4
Thanked 0 times in 0 posts
PostFri Nov 20, 2009 12:07 pm
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: ?

http://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
Back to top
View user's profile Send private message
palvarez
Really wants to Beat The GMAT!

Default Avatar

Joined: 24 Oct 2009
Posts: 199

Thanks given: 3
Thanked 19 times in 17 posts

GMAT Score: 710
PostFri Nov 20, 2009 1:09 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: ?

http://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.
Back to top
View user's profile Send private message
Reply from GMAT/MBA Admissions Expert
See All Featured Experts
Stuart Kovinsky
GMAT Instructor

Most Active Expert Most Thanked Expert
Joined: 08 Jan 2008
Posts: 2263

Thanks given: 0
Thanked 541 times in 467 posts
Location: Toronto

GMAT Score: 800
PostFri Nov 20, 2009 5:37 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: ?

http://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.)
_________________
Stuart Kovinsky, B.A. LL.B.
Academic Manager
Admissions Consultant
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST

GMAT Blogs

Learn more about me
Back to top
View user's profile Send private message
tgf
Just gettin' started!

Default Avatar

Joined: 15 Jul 2009
Posts: 16

Thanks given: 4
Thanked 0 times in 0 posts
PostSat Nov 21, 2009 6:36 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
Back to top
View user's profile Send private message
tgf
Just gettin' started!

Default Avatar

Joined: 15 Jul 2009
Posts: 16

Thanks given: 4
Thanked 0 times in 0 posts
PostSat Nov 21, 2009 7:22 am
Just one last question:

Whether the 5 donuts are identical or different is completely irrelevant when is it comes to the answer?
Back to top
View user's profile Send private message
Reply from GMAT/MBA Admissions Expert
See All Featured Experts
Stuart Kovinsky
GMAT Instructor

Most Active Expert Most Thanked Expert
Joined: 08 Jan 2008
Posts: 2263

Thanks given: 0
Thanked 541 times in 467 posts
Location: Toronto

GMAT Score: 800
PostSat Nov 21, 2009 8:10 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.
_________________
Stuart Kovinsky, B.A. LL.B.
Academic Manager
Admissions Consultant
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST

GMAT Blogs

Learn more about me
Back to top
View user's profile Send private message
Thanked by: tgf
Pooja Bhula
Just gettin' started!

Default Avatar

Joined: 24 Nov 2009
Posts: 29

Thanks given: 0
Thanked 0 times in 0 posts
PostWed Nov 25, 2009 4:43 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...
Back to top
View user's profile Send private message
Display posts from previous:   

Post New Topic   Post Reply All times are GMT - 7 Hours
Page 1 of 1
 
Forum Index GMAT Math Problem Solving
Most Active Members in Last 30 Days
1. harsh.champ 621 posts
2. shashank.ism 430 posts
3. ajith 342 posts
4. thephoenix 337 posts
5. money9111 332 posts
See More Top Beat The GMAT Members
Most Active Experts in Last 30 Days
1. lunarpower
Manhattan GMAT Teacher 		85 posts
2. Stuart Kovinsky
Kaplan GMAT Teacher 		71 posts
3. Lisa Anderson
Stacy Blackman Consulting 		50 posts
4. Testluv
Kaplan GMAT Teacher 		50 posts
5. Bryant@VeritasPrep
Veritas Prep 		30 posts
See More Top Beat The GMAT Experts