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

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

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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N+r-1 C r-1

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

Or n+r-1 C r-1
N=8
R=3
So ans = 7C2

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I do not get your approach, could you please explain what you mean by separators?
Thank you very much!
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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.
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ShankSouljaBoi
N=5 not 8

Halle | Julia | Drew

Number of ways = 7C2 = 21

IMO C

Bunuel can you explain this in a simpler manner please.

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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Bunuel please explain how to go about this!
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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
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Isn't 5 + 10 + 10 + 5 + 1 = 31 ?
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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.

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
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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
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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.

Bunuel can you please explain the logic behind the formula N+r-1CR-1??

How come we can’t say that the first donut has 3 choices it can go to and the second one also has 3 and so on. So we get 3^5

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