Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 16 Feb 2011
Posts: 195
Schools: ABCD

In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
21 Oct 2012, 16:52
1
This post received KUDOS
10
This post was BOOKMARKED
Question Stats:
49% (01:05) correct 51% (01:15) wrong based on 190 sessions
HideShow timer Statistics
In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5! Method1 (Long method): N= the child doesn't get anything 5NNN : 4!/3! * 5C5 = 4 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60 32NN : 4!/2! * 5C3*2C2 = 12*10 =120 311N : 4!/2! * 5C3*2C2 = 12*10 = 100 221N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180 2111 : 4!/3! * 5C2*3C3 = 4*10=40 If I add these numbers, it doesn't equal to 1024. What's my mistake? Thanks
Official Answer and Stats are available only to registered users. Register/ Login.



Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
22 Oct 2012, 07:29
2
This post received KUDOS
voodoochild wrote: In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!
Method1 (Long method):
N= the child doesn't get anything
5NNN : 4!/3! * 5C5 = 4 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60 32NN : 4!/2! * 5C3*2C2 = 12*10 =120 311N : 4!/2! * 5C3*2C2 = 12*10 = 100 221N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180 2111 : 4!/3! * 5C2*3C3 = 4*10=40
If I add these numbers, it doesn't equal to 1024. What's my mistake?
Thanks 5NNN : 4!/3! * 5C5 = 4  OK 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60  OK, divide by 2! because the two zero's (N) are indistinguishable 32NN : 4!/2! * 5C3*2C2 = 12*10 =120  OK 311N : 4!/2! * 5C3*2C2 = 12*10 = 100  it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct 221N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180  NO  it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts 2111 : 4!/3! * 5C2*3C3 = 4*10=40  NO  it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!. The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024. But why sweat all the way along????
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Manager
Joined: 16 Feb 2011
Posts: 195
Schools: ABCD

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
22 Oct 2012, 08:10
EvaJager wrote: But why sweat all the way along???? 4 word answer : "No pain, no gain"  something that my physical trainer taught me a few years ago.



Manager
Joined: 12 Oct 2011
Posts: 127
GMAT 1: 700 Q48 V37 GMAT 2: 720 Q48 V40

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
31 Oct 2012, 11:16
EvaJager wrote: voodoochild wrote: In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!
Method1 (Long method):
N= the child doesn't get anything
5NNN : 4!/3! * 5C5 = 4 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60 32NN : 4!/2! * 5C3*2C2 = 12*10 =120 311N : 4!/2! * 5C3*2C2 = 12*10 = 100 221N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180 2111 : 4!/3! * 5C2*3C3 = 4*10=40
If I add these numbers, it doesn't equal to 1024. What's my mistake?
Thanks 5NNN : 4!/3! * 5C5 = 4  OK 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60  OK, divide by 2! because the two zero's (N) are indistinguishable 32NN : 4!/2! * 5C3*2C2 = 12*10 =120  OK 311N : 4!/2! * 5C3*2C2 = 12*10 = 100  it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct 221N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180  NO  it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts 2111 : 4!/3! * 5C2*3C3 = 4*10=40  NO  it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!. The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024. But why sweat all the way along???? Why do you have to divide by 2! on 221N but don't have to divide by 2! on 311N?



Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
31 Oct 2012, 11:48
1
This post received KUDOS
BN1989 wrote: EvaJager wrote: voodoochild wrote: In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) (A) 4^5 (B) 5^4 (C) 5! (D) 4! (E) 4!*5!
Method1 (Long method):
N= the child doesn't get anything
5NNN : 4!/3! * 5C5 = 4 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60 32NN : 4!/2! * 5C3*2C2 = 12*10 =120 311N : 4!/2! * 5C3*2C2 = 12*10 = 100 221N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180 2111 : 4!/3! * 5C2*3C3 = 4*10=40
If I add these numbers, it doesn't equal to 1024. What's my mistake?
Thanks 5NNN : 4!/3! * 5C5 = 4  OK 41NN : 4!/2! * 5C4 *1C1 = 5*12 = 60  OK, divide by 2! because the two zero's (N) are indistinguishable 32NN : 4!/2! * 5C3*2C2 = 12*10 =120  OK 311N : 4!/2! * 5C3*2C2 = 12*10 = 100  it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct221N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 22) = 12 * 15 = 180  NO  it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts2111 : 4!/3! * 5C2*3C3 = 4*10=40  NO  it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!. The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024. But why sweat all the way along???? Why do you have to divide by 2! on 221N but don't have to divide by 2! on 311N? In the case of 221N we divide by 2! because we choose 2 groups of 2 one after the other and order when choosing them doesn't matter. Anyway, we permute them latter (see the factor of 4!). Similarly, in the case of 311N we should divide by 2! if we choose one single ball after the other 4! * 5C3 * 2C1 * 1C1/2!. If it is clear that we split the two remaining balls after choosing 3, we don't need the 2C1 and 1C1 factors. That's why 4! * 5C3.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Manager
Joined: 12 Oct 2011
Posts: 127
GMAT 1: 700 Q48 V37 GMAT 2: 720 Q48 V40

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
01 Nov 2012, 08:08
Thanks for your reply Eva, but I still haven’t fully understood it yet. Let’s use the 311N example, let’s say the candies are named A, B, C, D and E: One possible outcome would be ABCDEN, for this outcome there are 4! possible ways to distribute this outcome among the four people and 5C3 ways for the one person to get 3 candies so there are 240 possible arrangements of this kind. Now I don’t understand exactly how this would look like if we choose one candy after the other, so that this formula applies: 4! * 5C3 * 2C1 * 1C1/2! If you draw again ABC for the person who gets 3, you have D and E left for the next person so two choices and then you have again two choices about who out of the remaining two people gets the last candy?! So after ABC was picked you have 2C1*2!=4 options and you could permute this outcome in 4! ways, which obviously is wrong. How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?



Director
Joined: 22 Mar 2011
Posts: 608
WE: Science (Education)

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
01 Nov 2012, 09:03
BN1989 wrote: Thanks for your reply Eva, but I still haven’t fully understood it yet. Let’s use the 311N example, let’s say the candies are named A, B, C, D and E: One possible outcome would be ABCDEN, for this outcome there are 4! possible ways to distribute this outcome among the four people and 5C3 ways for the one person to get 3 candies so there are 240 possible arrangements of this kind. Now I don’t understand exactly how this would look like if we choose one candy after the other, so that this formula applies: 4! * 5C3 * 2C1 * 1C1/2! If you draw again ABC for the person who gets 3, you have D and E left for the next person so two choices and then you have again two choices about who out of the remaining two people gets the last candy?! So after ABC was picked you have 2C1*2!=4 options and you could permute this outcome in 4! ways, which obviously is wrong. How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!? How exactly does the process look like if I draw according to 4! * 5C3 * 2C1 * 1C1/2!?Think of carrying out the process in two steps: 1) Divide the 5 candies into 3 groups  3, 1, 1 Decide on the 3 candies somebody will get them  5C3  for example A, B, C (but think of all the possibilities) Which one is the first 1 single candy  2C1  D or E Which one is the second 1 single candy  1C1  E or D This would give 5C3 * 2C1 * 1C1 Divide by 2! because we are going to permute the groups of 3, 1, 1, 0, so we don't care about which single candy was chosen first. ABC, D, E is the same as ABC, E, D  what matters is that A, B, C are together, and D and E are single 2) After we have our groups of 3, 1, 1, 0 candies, we consider them as 4 distinct objects and just permute them  4!  because we have 4 different kids.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Manager
Joined: 12 Feb 2012
Posts: 130

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
02 Sep 2013, 14:12
I am getting 8!/(3!*5!)=56. I am using the stars bars method. (n+k1)C(k1). Why is this answer incorrect?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7938
Location: Pune, India

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
02 Sep 2013, 20:40
alphabeta1234 wrote: I am getting 8!/(3!*5!)=56. I am using the stars bars method. (n+k1)C(k1). Why is this answer incorrect? There are various methods of solving P&C questions and there are many formulas. You need to know the exact situation in which you can use a particular formula or better yet, you should understand the reason a particular method works in a particular situation. Stars bars method (or the formula (n+k1)C(k1)) is useful only when the objects to be distributed are identical. You split identical objects into different groups. You divide by 5! because the objects are identical. Here you can not use that method because the candies are different. Check out these posts on when to use which method: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 14 Aug 2012
Posts: 20

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
02 Sep 2013, 23:17
6
This post received KUDOS
there are total 5 candies. Lets say A,B,C,D and E. Candy A can be distributed among any 4 children. So there are 4 ways of distributing candy A. similarly candy B,C,D and E can be distributed in 4 ways.
So total ways of distribution is 4*4*4*4*4 = 4^5.
A



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7938
Location: Pune, India

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
05 Sep 2013, 20:37
VeritasPrepKarishma wrote: There are various methods of solving P&C questions and there are many formulas. You need to know the exact situation in which you can use a particular formula or better yet, you should understand the reason a particular method works in a particular situation. Stars bars method (or the formula (n+k1)C(k1)) is useful only when the objects to be distributed are identical. You split identical objects into different groups. You divide by 5! because the objects are identical. Here you can not use that method because the candies are different. Check out these posts on when to use which method: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/alphabeta1234 wrote: Why is the same formula (n+r1)C(r1) used for number of ways of dividing n identical items among r persons whom can receive 0,1,2 or more items . Is the formula used even if each person can receive 2 or 3 or any item? Why isn't that being factored in to the formula?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 1 mango?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 2 mango?
How many ways are there to distribute 10 mangoes among 4 kids if each kid must receive at least 3 mango?
Are you saying this formula gives us the same answer for all these questions? (10+41)C(41)? Certainly not. You will not have the same answer in every case. You can use the same formula but the value of n will vary in each case. Say if every kid must receive at least one mango, you take any 4 mangoes from the bunch of 10 (they are identical) and give one to each of the 4 kids. You cannot do this in more than 1 way because all the mangoes are the same and all the kids are receiving exactly one mango. Next, you are left with 6 mangoes and 4 kids and you need to distribute these 6 among the kids such that a kid may get no mango and another may get all etc. (You can do this because all the kids have already got a mango each and hence our condition is already satisfied.) This boils down to our previous question except that the value of n = 6 now, not n = 10. Now you use the formula as (6 + 4  1)C(4  1) = 9!/6!*3! You can do the same thing for at least 2 mangoes too.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 26 Jul 2015
Posts: 20

In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
09 Aug 2015, 14:14
Why do we ignore the case in which no candy is given to any child? It is literally in the question statement that the children could get zero candies: Quote: In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) Admittedly, I'm not a native English speaker but to me it's clear that I do not have to give a candy to any child. I can keep all candies to myself. Nor does the question state that all five candies must be given out to children. I could give out 0, 1, 2, 3, 4, or all five. So, using the awesome approach posted by shrikar23, we would get 5^5 because there are 5 ways of giving out each distinct candy: give it to no child, or to child 1, or to child 2 and so forth.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7938
Location: Pune, India

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
09 Aug 2015, 20:17
jhabib wrote: Why do we ignore the case in which no candy is given to any child? It is literally in the question statement that the children could get zero candies: Quote: In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy) Admittedly, I'm not a native English speaker but to me it's clear that I do not have to give a candy to any child. I can keep all candies to myself. Nor does the question state that all five candies must be given out to children. I could give out 0, 1, 2, 3, 4, or all five. So, using the awesome approach posted by shrikar23, we would get 5^5 because there are 5 ways of giving out each distinct candy: give it to no child, or to child 1, or to child 2 and so forth. The question states that you need to "distribute 5 candies". So all 5 candies must be given out. It is possible that one or more children may not get any candy and one or more may get more than one candies but all 5 have to be distributed.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



NonHuman User
Joined: 09 Sep 2013
Posts: 13827

Re: In how many ways can 5 different candiesbe distributed among [#permalink]
Show Tags
04 Oct 2017, 02:10
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: In how many ways can 5 different candiesbe distributed among
[#permalink]
04 Oct 2017, 02:10






