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

 It is currently 14 Nov 2018, 01:26

### 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

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

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### \$450 Tuition Credit & Official CAT Packs FREE

November 15, 2018

November 15, 2018

10:00 PM MST

11:00 PM MST

EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth \$100 with the 3 Month Pack (\$299)
• ### Free GMAT Strategy Webinar

November 17, 2018

November 17, 2018

07:00 AM PST

09:00 AM PST

Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 13 May 2009
Posts: 18
Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

### Show Tags

07 Aug 2009, 05:57
8
00:00

Difficulty:

95% (hard)

Question Stats:

36% (02:15) correct 64% (02:25) wrong based on 162 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
Manager
Joined: 18 Jul 2009
Posts: 164
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

07 Aug 2009, 06:51
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

##### General Discussion
Manager
Joined: 17 Jul 2009
Posts: 242
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

07 Aug 2009, 06:59
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
Joined: 22 Jul 2009
Posts: 168
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

### Show Tags

22 Sep 2009, 15:07
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
Joined: 11 Sep 2009
Posts: 129
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

### Show Tags

22 Sep 2009, 15:36
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
Joined: 07 Mar 2011
Posts: 44
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

### Show Tags

19 Jul 2011, 05:35
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
Status: ==GMAT Ninja==
Joined: 08 Jan 2011
Posts: 198
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

19 Jul 2011, 06:01
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.

Director
Joined: 14 Dec 2017
Posts: 508
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

### Show Tags

10 Jun 2018, 23:30
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

Thanks,
GyM
_________________
Manager
Joined: 05 Jan 2014
Posts: 61
Location: India
Schools: IIM (S)
GMAT 1: 610 Q47 V26
GPA: 3.76
WE: Information Technology (Computer Software)
Re: Larry, Michael, and Doug have five donuts to share. If any one of the  [#permalink]

### Show Tags

02 Jul 2018, 19:20
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, 19:20
Display posts from previous: Sort by

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

 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®.