BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Donut combination

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 154
Joined: Tue Aug 26, 2008 12:59 pm
Location: Canada
Thanked: 4 times

Re: Donut combination

by canuckclint » Wed Sep 03, 2008 11:11 am
Ian Stewart wrote:
sudhir3127 wrote:
i would go with 21 as well..

formula is

"distributing n identical things among r persons such that any person might get any number ( incl 0) is n+r -1 C r-1"
I doubt you'll ever need this formula on the GMAT, but if you're interested in how to do the problem (for any number of donuts) without a formula, or if you want to know why the formula works:

Line up the five donuts:

O O O O O

All we want to do is partition the donuts among three people. For example, we could split up the donuts as follows:

O | O O O | O

That is, we could give the first person one donut, the second person three donuts, and the third person one donut. Or we could partition like this:

| O O O O | O

That is, the first person gets zero donuts, the second gets four, and the third gets one. Each different partition assigns different numbers of donuts to the group of people. If you count, we have seven slots in which we could place the two partition markers (the '|' symbols), and the order in which we place them does not matter: 7C2. You can generalize the approach to get the formula Sudhir mentions above.
--
This is an incredible and clear approach by Ian. Thanks for the symbols!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 154
Joined: Tue Aug 26, 2008 12:59 pm
Location: Canada
Thanked: 4 times

Re: Donut combination

by canuckclint » Wed Sep 03, 2008 11:25 am
4meonly wrote:
Ian Stewart wrote: | O O O O | O
If you count, we have seven slots in which we could place the two partition markers (the '|' symbols), and the order in which we place them does not matter:
I can find only 6 places...


1 O 2 O 3 O 4 O 5 O 6

where is 7th?

please, help me
You need to think of these in terms of alphabets

So how many different ways are there to put aa and 5 d's.

aaddddd

7!/(2! * 5!). You divide the repeaters of double alphabets. equivalent to (7 c 2).
aacc will be 4!/(2! * 2!) etc.

Another way to do think is to have 7 slots and put 2 a's in there.
(7 c 2). Or 5 c's in 7 slots (7 c 5) equivalent to (7 c 2)
(n c r) equivalent to (n c (n-r)) always.

Cheers,
Clinton
Math Minor, Comnputer Science major
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member
Thanked: 27 times
GMAT Score:660(

by 4meonly » Wed Sep 03, 2008 11:35 am
Thank you

So how many different ways are there to put aa and 5 d's.
aaddddd
7!/(2! * 5!). You divide the repeaters of double alphabets. equivalent to (7 c 2).
aacc will be 4!/(2! * 2!) etc.

I am OK with this



Another way to do think is to have 7 slots and put 2 a's in there.
(7 c 2). Or 5 c's in 7 slots (7 c 5) equivalent to (7 c 2)
(n c r) equivalent to (n c (n-r)) always.

This is my problem. I really dont see 7 slots... I see 6 :(

I've met such things in questions and solutions but really i do not understand :-(
where is the 7th?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

by Ian Stewart » Wed Sep 03, 2008 2:54 pm
4meonly,

CanuckClinton explained this in a different way, but it's because you can put two of the partitions next to each other that you have seven slots, and not six.

We can think of it this way: first place one partition, then place the next. In the end we'll need to divide the number of ways we can do this by two, however, because the order of the partitions doesn't matter.

So, if we only insert one partition, as you correctly say we have exactly six choices. We might do as follows:

O O O O | O

Now, when we insert the second partition, we have seven choices- seven because we could put the second partition immediately to the left, or immediately to the right, of the first (of course the result is the same either way, which is why we divide by 2; the order of the partitions doesn't actually matter). That is, we could put the second partition so the diagram looks like this:

O O O O | | O

And with this partition, Larry gets 4, Michael gets none, and Doug gets 1. So we had six choices for where to place the first partition, then seven for where to place the second (including both spots on either side of the first partition), and finally we divide 7*6 by 2. Alternatively, you are choosing 2 slots from 7 where order does not matter, so you have 7C2 choices.

A bit of a long day today, so I hope the above is clear!
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member
Thanked: 27 times
GMAT Score:660(

by 4meonly » Thu Sep 04, 2008 7:33 am
Ian Stewart wrote: O O O O | | O

And with this partition, Larry gets 4, Michael gets none, and Doug gets 1. So we had six choices for where to place the first partition, then seven for where to place the second (including both spots on either side of the first partition), and finally we divide 7*6 by 2. Alternatively, you are choosing 2 slots from 7 where order does not matter, so you have 7C2 choices.
I last I've got it!!!! ))))
Thank you!

:D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply