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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  [#permalink]

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New post 01 Nov 2019, 04:04
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 01 Nov 2019, 04:24
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Bunuel wrote:
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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total possible ways to distribute donuts as whole no
5c1+5c2+5c3+5c4+5c5 = 5+10+10+5+1 ; 21
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 01 Nov 2019, 12:00
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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Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 01 Nov 2019, 12:02
Or n+r-1 C r-1
N=8
R=3
So ans = 7C2

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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 09 Nov 2019, 23:51
ShankSouljaBoi wrote:
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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I do not get your approach, could you please explain what you mean by separators?
Thank you very much!
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Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post Updated on: 26 Dec 2019, 04:12
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1
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.

If this was helpful, kindly give Kudo.

Originally posted by ShankSouljaBoi on 10 Nov 2019, 05:51.
Last edited by ShankSouljaBoi on 26 Dec 2019, 04:12, edited 2 times in total.
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 16 Nov 2019, 06:09
ShankSouljaBoi
N=5 not 8
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 25 Feb 2020, 21:14
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1
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
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 27 Mar 2020, 23:30
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Bunuel wrote:
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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Halle | Julia | Drew

Number of ways = 7C2 = 21

IMO C
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi  [#permalink]

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New post 07 Apr 2020, 04:41
Bunuel can you explain this in a simpler manner please.
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Re: Halle, Julia and Drew have 5 donuts to share. If one of them can be gi   [#permalink] 07 Apr 2020, 04:41
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