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

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.

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

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}

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.

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

Why 3^5 is not a correct answer?
Considering each donut has 3 possibilities(L,M &D)...
Kindly regards

Because the donuts are not distinct.

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

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)

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

\(0+1+4\) - 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 = \(6+3+3+6+3 = 21.\)

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

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

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

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

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.

Hi Bunuel, what am I missing here? My answer is 6C2. Choose 2 slots out of 6

D= donut ; Slot = S

S1 D S2 D S3 D S4 D S5 D S6

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