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In how many ways can 5 different fruits be distributed among
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itsworththepain
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Joined: 07 Jun 2014
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In how many ways can 5 different fruits be distributed among [#permalink] New post  Updated on: 27 Jun 2014, 00:47
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A
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C
D
E

Difficulty:

55% (hard)

Question Stats:

53% (01:04) correct 47% (01:27) wrong based on 311 sessions

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In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!

Show: ::
Correct Answer is 1024 and I understand the reasoning behind that.


But the approach I followed was:


Say we keep the 5th fruit aside and distribute the other 4 fruits among the 4 children
No of ways to distribute 4 fruits among 4 children = 4!

Now for each of these 4! Combinations, 5th fruit can be distributed to any of the 4 children
i.e. 4 new combinations for each of the 4! combinations
No of ways to distribute 5th fruit = 4*4!

5th fruit can be selected in 5 different ways
Total combinations are 5*4*4! = 480

What’s wrong here? What is it that I am missing here? What are the other 1024-480 combinations that I am missing?
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Originally posted by itsworththepain on 26 Jun 2014, 22:53.
Last edited by Bunuel on 27 Jun 2014, 00:47, edited 2 times in total.
Edited the question and added the OA.
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VeritasKarishma
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Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  26 Jun 2014, 23:08
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sagarsameer wrote:
Question: In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!

Correct Answer is 1024 and I understand the reasoning behind that.


But the approach I followed was:


Say we keep the 5th fruit aside and distribute the other 4 fruits among the 4 children
No of ways to distribute 4 fruits among 4 children = 4!

Now for each of these 4! Combinations, 5th fruit can be distributed to any of the 4 children
i.e. 4 new combinations for each of the 4! combinations
No of ways to distribute 5th fruit = 4*4!

5th fruit can be selected in 5 different ways
Total combinations are 5*4*4! = 480

What’s wrong here? What is it that I am missing here? What are the other 1024-480 combinations that I am missing?


I believe you picked up this question from my blog post so you know how to correctly solve it. So I will focus on only pointing out the error you committed. To start off, how do you get "No of ways to distribute 4 fruits among 4 children = 4! "? You are not given that each child gets only one fruit. If each child were to receive only one fruit, then you could have said that the first fruit can be given away in 4 ways, 2nd fruit in 3 ways and so on to get 4*3*2*1.

Some children may get none so others may get 2 or 3 or 4. Each fruit can be given out in 4 ways so 4*4*4*4*4 = 1024
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itsworththepain
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Joined: 07 Jun 2014
Posts: 14
Location: India
Schools: Wharton '17 (D), Anderson FEMBA '17 (S), Duke Fuqua (II), Stern '17 (S)
GMAT 1: 720 Q49 V38
GPA: 2.91
WE:Consulting (Energy and Utilities)
Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  26 Jun 2014, 23:12
VeritasPrepKarishma wrote:
sagarsameer wrote:
Question: In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!

Correct Answer is 1024 and I understand the reasoning behind that.


But the approach I followed was:


Say we keep the 5th fruit aside and distribute the other 4 fruits among the 4 children
No of ways to distribute 4 fruits among 4 children = 4!

Now for each of these 4! Combinations, 5th fruit can be distributed to any of the 4 children
i.e. 4 new combinations for each of the 4! combinations
No of ways to distribute 5th fruit = 4*4!

5th fruit can be selected in 5 different ways
Total combinations are 5*4*4! = 480

What’s wrong here? What is it that I am missing here? What are the other 1024-480 combinations that I am missing?


I believe you picked up this question from my blog post so you know how to correctly solve it. So I will focus on only pointing out the error you committed. To start off, how do you get "No of ways to distribute 4 fruits among 4 children = 4! "? You are not given that each child gets only one fruit. If each child were to receive only one fruit, then you could have said that the first fruit can be given away in 4 ways, 2nd fruit in 3 ways and so on to get 4*3*2*1.

Some children may get none so others may get 2 or 3 or 4. Each fruit can be given out in 4 ways so 4*4*4*4*4 = 1024



Ahh... I see the issue now, thanks a lot!! And yes indeed the article was from your post... Thanks again!!
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Bunuel
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Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  27 Jun 2014, 00:48
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sagarsameer wrote:
In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!

Show: ::
Correct Answer is 1024 and I understand the reasoning behind that.


But the approach I followed was:


Say we keep the 5th fruit aside and distribute the other 4 fruits among the 4 children
No of ways to distribute 4 fruits among 4 children = 4!

Now for each of these 4! Combinations, 5th fruit can be distributed to any of the 4 children
i.e. 4 new combinations for each of the 4! combinations
No of ways to distribute 5th fruit = 4*4!

5th fruit can be selected in 5 different ways
Total combinations are 5*4*4! = 480

What’s wrong here? What is it that I am missing here? What are the other 1024-480 combinations that I am missing?


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in-how-many-ways-can-5-different-candies-be-distributed-in-141072.html
in-how-many-ways-can-5-different-candiesbe-distributed-among-141070.html
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Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  05 Mar 2016, 14:39
itsworththepain wrote:
In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!




we can eliminate right away C and D.
E - 4!*5!=120*24 = 240*12=480*6=960x3 = 2880...way too much.. so out.
B - 5^4 - well, we have 5 fruits and 4 children..each children can be given 1, 2, 3, 4, or 5 fruits. so most likely should be 4^5 than 5^4.
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Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  30 Jul 2019, 08:54
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itsworththepain wrote:
In how many ways can 5 different fruits be distributed among four children? (Some children may get more than one fruit and some may get no fruits.)

(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!



Since each child can get 0 to 5 fruits, the number of ways the 5 different fruits be distributed among four children is 4^5.

Answer: A
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In how many ways can 5 different fruits be distributed among [#permalink] New post  01 Aug 2019, 06:22
I was trying to use the stick method (that looks like this A B | C | D | E) but that didn't work. I suppose it's because the fruits are different not identical?
I could solve the question with the basic 5 slots 4 choices each approach, but I'm trying to understand the stick method and cases where it's applicable. Could someone elaborate on why the stick method didn't work in this case. Does it only work for identical-object cases? Am I missing something obvious here? Thanks
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Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  11 Sep 2020, 00:18
Hey , thanks all for great post.
I have one query , i am trying to understand the difference between (4 raise to power 5 ) thinking, one fruit can be given to any of 4 children--> 4*4*4*4*4

Also ,(5 raise to power 4) wherein each children can take any of 5 available fruits..

Can anyone please explain the difference..
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Re: In how many ways can 5 different fruits be distributed among [#permalink] New post  19 Sep 2020, 08:38
CrackVerbalGMAT, why cannot I use the below formula for this question?

(n+r−1) C (r−1)

Is it because objects are not identical?
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