Halle, Julia and Drew have 5 donuts to share. If one of them can be gi

Question

Halle, Julia and Drew have 5 donuts to share. If one of them can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed.

A. 19
B. 20
C. 21
D. 23
E. 37

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Raksat · Accepted Answer

The possible combinations possible 
0-2-3 : No of ways =6
0-1-4 : No of ways = 6
1-2-2 : No of ways = 3
0-0-5 : No of ways = 3
1-1-3 : No of ways = 3
Hence the total no of ways =21

ShankSouljaBoi · Answer

A|B|C 5 identical donuts 
2 separators and 5 donuts = 7!
Separators are identical and so are donuts.
Hence ans = 7!/(5!*2!) = 21

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Archit3110 · Answer

N+r-1 C r-1

(5+3-1)C (3-1)
7c2 ; 21
Option C

ShankSouljaBoi · Answer

Use this, its basically the same thing
n+r-1 C r-1
N=5
R=3
So ans = 7C2

Now coming to your query.

How many people are there ? Three, who ? A B and C. Are these 3 people identical? Not possible !
How many donuts have ti be distributed ? 5 . Are these identical ? Yes . Heck, over counting will happen. Just keep this thought in mind and lets move forward.

For explaining to u in a better manner lets assume A B C be three bags and not people. 
Further lets assume i am Donut 1 . Cool So mr soulja where you gonna jump ? I say, any one.
Same with other 4 Mr. Soulja's.

...
What this means is any donut can go in any bag , right ?
Bag A can have all 5 or even Zero.

So when such a distribution problem comes you can apply the illustrated method.

Now lets see what i exactly did there.

Lets introduce two sticks or separators or in our explanation i would like to call them partitions.
These partitions are a deciding factor.
How ?

A partition 1 B partition 2 C --- this combination reprsents any number of donuts in any bag. But at least one in each bag.
partition 1 partition 2 ABC ---- this combination means No donuts to anyone imagine this OOOOOABC
partition 1 A Partition2 BC --- this combination means at least one is before partition 1 , at least one is in A, and remaining can be distributed anywhere.

So basically this is a permutation of 7 items 2 idenctical partitions and 5 exact (donuts or Mr soulja's)

Hence, 7!/(5!2!)

Ill be happy to entertain further questions.

Avysar · Answer

Though we can use the formula (n+r-1)C(r-1) as the shortcut, we can solve this logically, if we understand the possible arrangements
First, we can use the property that n items can be arranged in n! Ways
Second, if out of n items there are x repetitions then the number of arrangements would be n!/x!
Coming to the question, there are 3 guys and 5 doughnuts. So how all the number 5 can be divided among the three...
If 0,0&5 then 3 ways because it is 3!/2!
If1,1,3 then again 3!/2!=3 ways
If 2,2,1 agan 3!/2!= 3 ways
If 0,1,4 then 3!
If 0,2,3 then 3!
No other arrangement will add upto 5.
So, in all, 3+3+3+6+6=21
Hence C

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yashikaaggarwal · Answer

Halle Julia Drew No. of ways
5 0 0 3C2 = 3
4 1 0 3C1 = 6
3 2 0 3C1 = 6
3 1 1 3C2 = 3
2 2 1 3C2 = 3

here let's suppose Halle got all 5 donuts giving away 0 donut to rest two. 
plus all three have equal chance of getting this 5 donuts. so it can be 5,0,0 or 0,5,0 or 0,0,5
therefore 3 different ways. 
similarly with 4,1,0 
this can be written as 4,1,0 or 4,0,1
1,0,4 or 1,4,0
0,4,1 or 0,1,4 = 6 ways 
Similarly you can calculate rest, hence 3+6+6+3+3 = 21
 SiffyB  if you further have query ask away

SiffyB · Answer

Isn't  5 + 10 + 10 + 5 + 1 = 31  ?

TarunKumar1234 · Answer

Distributing 5 identical objects in 3 non-identical bins. (n+r-1) C (r-1)= (5+3-1) C (3-1) = 7C2 = 21. So, Ans C.

akadiyan · Answer

Halle, Julia and Drew have 5 donuts to share. If one of them can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed.

Number of items = n = 5
Number of people = r = 3

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0, 1, 2 or more items  n+r−1Cr−1

5+(3-1)C(3-1) = 7C2 = 21

Ans: C

CrackverbalGMAT · Answer

You are using the equation of  2^n - 1 , which is the total number of ways of choosing from n distinct things, where choosing nothing is a possibility.

Here the question is the number of ways of distributing n similar things among a certain number of people, where 0 is also a possibility.

Hope this helps

Arun Kumar

msanmartin · Answer

Its pretty simple, we could just used the combinatorial technique known as  Stars and Bars .

We have 5 donuts (5 stars):

* * * * *

These donuts must be divided into three groups hence we separate them using two bars

A B C
* | * * | * *

Each division representing the number of donuts each person gets. (In this example A gest 1 B gets 2 and C gets 2 as well).

Since we have 7 objetects 5 of them are the same and the other two are the same, we just operate 7C2 which is 21.

hjharsh · Answer

Can someone please help me explain:

Can we not look at it this way?

Donut 1 : 3 options
Donut 2 : 3 options

likewise 3x3x3x3x3 = 243

Bunuel · Answer

Here, donuts are assumed to be identical, so this method does not apply.

hjharsh · Answer

I understand that donuts are identical - but for each donut - there are three options - Halle, Julie, Drew - does not work

Also I am getting confused whether to use n+r-1 C r-1 method or the one you showed simply putting 2 bars in between - I am asking more generally for all such questions

Krunaal · Answer

It does not work because when the 5 donuts are identical, it does not matter which donut went to whom. For eg. if the 5 donuts were different i.e. A B C D E, then Halle getting {A and B} or {C and D} - both cases would be counted, but when donuts are identical i.e. D D D D D, it will only be one case of Halle getting two donuts {D and D} Here's theory on P&C that will expose you to different cases, and help you understand when to apply which method: https://gmatclub.com/forum/combinatoric ... 06266.html

Halle, Julia and Drew have 5 donuts to share. If one of them can be gi

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Halle, Julia and Drew have 5 donuts to share. If one of them can be gi

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