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

Question

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

Bunuel · Accepted Answer

Consider this: we have 5 donuts d and 2 separators | , like: ddddd|| . How many permutations (arrangements) of these symbols are possible? Total of 7 symbols (5+2=7), out of which 5 d 's and 2 | 's are identical, so \frac{7!}{5!2!}=21 . We'll get combinations like: dd|d|dd this would mean that Larry got 2 donuts, Michael got 1 donut and Doug got 2 donuts, so to the left of the first separator are Larry's donuts, between the separators are Michael's donuts and to the right of the second separator are Doug's donuts Answer: A. This can be done with direct formula as well: The total number of ways of dividing n identical items (5 donuts in our case) among r persons or objects (3 persons in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is {n+r-1}_C_{r-1} . In our case we'll get: {n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21 . Similar question: integers-less-than-85291.html#p710836

bhushan252 · Answer

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

AKProdigy87 · Answer

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.

pankgarg · Answer

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

GSBae · Answer

If you get confused about combinations, there's a simple way to count these combinations as well, by counting the number of ways 5 can be summed with 3 numbers.

{5,0,0} = 3 possibilities. 
{4,1,0} = 6 possibilities. 
{3,2,0} = 6 possibilities. 
{3,1,1} = 3 possibilities. 
 {2,2,1} = 3 possibilities.  
Total = 21 possibilities.

Tip: For each set, we only have to consider numbers less than the first; for instance, we wouldn't consider {2,3,0} because that's already accounted for in a permutation of {3,2,0}

deya · Answer

Dear Bunuel,

Thanks for the great explanation. But there is a catch that is not clear to me. I read all other similar types of questions and also went through the explanations given by you. There you have used 3 separators for all the cases but here you have used 2. Could you tell me how would I know that how many separators should I use.

Thanks in advance. :-D

Bunuel · Answer

Distributing between 4 use 3 separators;
Distributing between 3 use 2 separators.

ankyrules · Answer

Keep in mind that 2 people might not even get any donuts at all.. so total 7! and 5 donuts are identical and 2 not given are identical .. so 7!/ 5! 2! = 21

kinjiGC · Answer

L + M + D = 5
As L,M,D >= 0

We need to distribute 5 donots and 2 empty vessels. So 7!/(5!*2!) or 7C2 = 21 - A)

ScottTargetTestPrep · Answer

Let D be a donut, so we have DDDDD to distribute among three people. Also, we can use “|” as a separator, and we need two of them since there are three people. For example, D|DDD|D means Larry gets 1 donut, Michael 3, and Doug 1, and DDDDD|| means Larry gets 5 donuts, Michael 0, and Doug 0. Thus, the problem becomes how many ways we can arrange 5 Ds and 2 strokes. To solve it, we can use the formula for permutation of indistinguishable objects:

7!/(5! x 2!) = (7 x 6 x 5!)/(5! x 2) = 42/2 = 21

Answer: A

TimeTraveller · Answer

0+1+4  - this can be done in  6  ways:  014, 104,401,140,410,041

1+1+3  - this can be done in  3  ways:  \frac{3!}{2!}

1+2+2  - this can be done in  3  ways:  \frac{3!}{2!}

2+3+0  - this can be done in  6  ways

0+0+5  - this can be done in  3  ways

Total no. of ways =  6+3+3+6+3 = 21.

GyMrAT · Answer

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

aniket16c · Answer

Consider 5 donuts: O O O O O

Divide it among 3 people: 
Case 1: L=0
1.1: L=0, M=6, D=0
1.2: L=0, M=5, D=1 ... 
Thus we will be able to do this till L=0, D=0, L=6. Total = 6 possibilities

Case 2: L=1
L=1,M=5, D=0.
Similar to above we can do this for Total = 5 possibilities

Thus, we can continue such iterations till L=6

From first two cases we can simply observe that there is a series formed: 6+5+4+3+2+1

6+5+4+3+2+1 =21

Rakhi08 · Answer

Hi math experts,

The question does not mention if donuts are identical or distinct. If GMAT questions are silent on this aspect, do we assume it to be identical?
Are there any other assumptions we take on questions on GMAT ?

Thanks in advance

Fdambro294 · Answer

Assuming each of the 5 donuts are Identical, we are distributing 5 identical items to 3 distinct “groups” of people (L, M, and D)

It does not matter which donuts each person gets - only the number of donuts matters in terms of producing a unique distribution.

Any person can receive 0 donuts or all the donuts. All 5 donuts will be distributed in each arrangement.

We can set up a Linear equation as:

L + M + D = 5

Where L , M, and D are all greater than or equal to >/= 0

We can look at the number of unique arrangements visually by placing Partitions between each Variable and drawing out 5 donuts:

* | * * | * *

In order to divide the identical donuts among the 3 people, we need 2 partitions

In the above visual representation:
L = 1 donut
M = 2 donuts 
D = 2 donuts

We can move the 2 partitions around to show all the unique possibilities, for example:

| | * * * * *

L = 0 ——- M = 0 ——- D = All 5

Or

* * | * * * |

L = 2—- M = 3 ——- D = 0

The number of unique distributions will be given by the number of way to Arrange and “shuffle around” the 7 elements: of which we have 2 identical petitions and 5 identical stars representing the donuts.

We can do this in: (7 c 2) ways or

7! / (2! * 5!) =

(7 * 6 ) / 2 =

21 unique distributions

Posted from my mobile device

IanStewart · Answer

You're completely right -- in real life, donuts often are distinct (a chocolate donut and a jelly donut are different), so there's no way to guess, reading this question, whether we're meant to think the donuts are identical or different, and it would be entirely reasonable to think they're different.

If they're identical, the problem becomes a standard partition problem, and the answer is 21 (as solved in many posts above), though I've never once seen a partition problem on the actual GMAT, so I don't think the intended meaning of this question is worth worrying about. If the donuts are different, we have 3 choices for each donut (give it to the first, second or third person), and the answer becomes 3^5. It's a poorly worded prep company question, and a real GMAT question would always be clear about whether the items are identical or different, so that's not an issue you'll need to worry about on the real test.

Sahith_Manikanta · Answer

Hi Brunel. I have a doubt here.
We're sharing 5 donuts among 3 individuals here right, and we don't have any constraint as to who will get any number of donuts.
So, each donut will have 3 options to go to (3 individuals). So, shouldn't the answer be 3x3x3x3x3 = 3^5.

The method you've stated is very valid, but my method is from questions like "3 letters need to be posted into 4 letter boxes". Where each letter has 4 options to go into, so 4^3. I don't seem to find any difference between these questions.

If you think the method I stated doesn't belong here pls clarify it.  GMATNinja   KarishmaB . Anyone pls clear my doubt.

KarishmaB · Answer

The two situations are different. You are talking about distributing distinct objects - 3 different letters. Here this question involves distributing 5 identical objects (items of a given food are considered identical unless stated otherwise. So 5 apples means identical apples). The handling of distinct objects vs identical objects is very very different. This question will be solves using separators as Bunuel mentioned above. I have discussed various such cases in my section on Distributions in Combinations study module. You can check it out on Sunday through Super Sundays program. Details here: https://youtu.be/gN_vlDpUflo

DrMock · Answer

@Bunuel When the question doesn't mention that they are distinct (in cases where they could be distinct or not depending on the scenario - like donuts, cars, chairs etc.), do we assume they aren't? That seems to be the case for this question.

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

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

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