GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Aug 2018, 05:27

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Larry, Michael, and Doug have five donuts to share. If any one of the

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
Joined: 13 May 2009
Posts: 19
Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 07 Aug 2009, 06:57
8
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

42% (01:21) correct 58% (02:08) wrong based on 156 sessions

HideShow timer Statistics

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
Most Helpful Community Reply
Manager
Manager
User avatar
Joined: 18 Jul 2009
Posts: 166
Location: India
Schools: South Asian B-schools
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 07 Aug 2009, 07:51
5
7
OA A

This is little trivial rule in combinations

The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r-1)C(r-1)

in this problem n=5 & r=3 hence 7C2 = 21 ways

**PS : if you want no of choices excluding 0 then => (n-1)C(r-1) = 4C2 = 6 ways
_________________

Bhushan S.
If you like my post....Consider it for Kudos :-D

General Discussion
Senior Manager
Senior Manager
avatar
Joined: 17 Jul 2009
Posts: 255
Concentration: Nonprofit, Strategy
GPA: 3.42
WE: Engineering (Computer Hardware)
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 07 Aug 2009, 07:59
question is not clear tho, if any one men can take from 0 to 5 donuts, is it allowed that together they only take less than 5? as in one guy take 1, the other take nothing. well that would make this a lil more complicated...
Manager
Manager
User avatar
Joined: 22 Jul 2009
Posts: 188
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 22 Sep 2009, 16:07
1
bhushan252 wrote:
OA A

The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r-1)C(r-1)


Good to know ... +1 for that

Btw, this is a Manhattan Gmat question. Manhattan's explanation is 1 page long.
_________________

Please kudos if my post helps.

Manager
Manager
avatar
Joined: 11 Sep 2009
Posts: 129
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 22 Sep 2009, 16:36
The other way to look at it is to simply have 5 donuts, with two dividers. You have 7 slots:

_ _ _ _ _ _ _

In those slots, you can either put a donut, O, or a divider, |. The number preceding the first divider goes to Person 1, the number between the two dividers goes to Person 2, and the number after the 2nd divider goes to Person 3.

For example:

O O | O O | O
| | O O O O O
| O | O O O O

etc.

You choose 2 spots out of the 7 to put the dividers, hence the answer is 7C2.
Intern
Intern
avatar
Joined: 07 Mar 2011
Posts: 44
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 19 Jul 2011, 06:35
1
Here is another intuitive way, but certainly the formula will help in bigger calculations.

to get the answer, see how can we get sum "5" with 3 numbers.

1) 0,0,5 = 3 combinations or 3! /2!

2) 0,1,4 = 6 combinations or 3!

similarly
3) 0,2,3 = 6 cobinations
4) 1,1,3 = 3 combination
5) 2,2,1 = 3 combination

total =21
Manager
Manager
User avatar
Status: ==GMAT Ninja==
Joined: 08 Jan 2011
Posts: 218
Schools: ISB, IIMA ,SP Jain , XLRI
WE 1: Aditya Birla Group (sales)
WE 2: Saint Gobain Group (sales)
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 19 Jul 2011, 07:01
bhushan252 wrote:
OA A

This is little trivial rule in combinations

The formula for distributing n identical things amongst r people such that any person might get any number of things ( including 0) is (n+r-1)C(r-1)

in this problem n=5 & r=3 hence 7C2 = 21 ways

**PS : if you want no of choices excluding 0 then => (n-1)C(r-1) = 4C2 = 6 ways


dear bhushan
got your formula for cases when assigning 0 is allowed

please explain the formula for cases where assigning of 0 is not allowed

for example the question below

If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible

is it as simple as it looks

here is my understanding
9 total digits out of which 3 places to assign
we cant assign 0

so the total no of ways will be (9-1) C (3-1)= 8 C 2 =28

is it correct
_________________

WarLocK
_____________________________________________________________________________
The War is oNNNNNNNNNNNNN for 720+
see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html
do not hesitate me giving kudos if you like my post. :)

Senior Manager
Senior Manager
User avatar
G
Joined: 14 Dec 2017
Posts: 459
CAT Tests
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 11 Jun 2018, 00:30
lbsgmat wrote:
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


Using the slot method, where D denotes the donut & l denotes the slot.

D D D D D l l

So # of ways of dividing the donuts is the # of possible arrangements of 5 D's & 2 l's = 7!/5!*2! = 21

Answer A.

Thanks,
GyM
Intern
Intern
avatar
B
Joined: 05 Jan 2014
Posts: 46
GMAT 1: 610 Q47 V26
WE: Information Technology (Computer Software)
GMAT ToolKit User CAT Tests
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

Show Tags

New post 02 Jul 2018, 20:20
Nowhere it is mentioned that donuts are identical. Shall we assume the same in such cases?
Re: Larry, Michael, and Doug have five donuts to share. If any one of the &nbs [#permalink] 02 Jul 2018, 20:20
Display posts from previous: Sort by

Larry, Michael, and Doug have five donuts to share. If any one of the

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.