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In how many ways can 5 different candiesbe distributed among

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In how many ways can 5 different candiesbe distributed among [#permalink] New post 21 Oct 2012, 16:52
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

27% (02:02) correct 72% (00:58) wrong based on 40 sessions
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

5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40

If I add these numbers, it doesn't equal to 1024. What's my mistake?


Thanks
[Reveal] Spoiler: OA
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 22 Oct 2012, 07:29
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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

5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40

If I add these numbers, it doesn't equal to 1024. What's my mistake?

Thanks


5-N-N-N : 4!/3! * 5C5 = 4 - OK
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK
3-1-1-N : 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
2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 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
2-1-1-1 : 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????
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 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. :)
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 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

5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40

If I add these numbers, it doesn't equal to 1024. What's my mistake?

Thanks


5-N-N-N : 4!/3! * 5C5 = 4 - OK
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK
3-1-1-N : 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
2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 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
2-1-1-1 : 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 2-2-1-N but don't have to divide by 2! on 3-1-1-N?
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 31 Oct 2012, 11:48
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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

5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40

If I add these numbers, it doesn't equal to 1024. What's my mistake?

Thanks


5-N-N-N : 4!/3! * 5C5 = 4 - OK
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK
3-1-1-N : 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
2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 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
2-1-1-1 : 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 2-2-1-N but don't have to divide by 2! on 3-1-1-N?


In the case of 2-2-1-N 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 3-1-1-N 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.
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 01 Nov 2012, 08:08
Thanks for your reply Eva, but I still haven’t fully understood it yet.
Let’s use the 3-1-1-N example, let’s say the candies are named A, B, C, D and E:
One possible outcome would be ABC-D-E-N, 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!?
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 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 3-1-1-N example, let’s say the candies are named A, B, C, D and E:
One possible outcome would be ABC-D-E-N, 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.
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 02 Sep 2013, 14:12
I am getting 8!/(3!*5!)=56. I am using the stars bars method. (n+k-1)C(k-1). Why is this answer incorrect?
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 02 Sep 2013, 20:40
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alphabeta1234 wrote:
I am getting 8!/(3!*5!)=56. I am using the stars bars method. (n+k-1)C(k-1). 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+k-1)C(k-1)) 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 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 02 Sep 2013, 23:17
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
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Re: In how many ways can 5 different candiesbe distributed among [#permalink] New post 05 Sep 2013, 20:37
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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+k-1)C(k-1)) 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 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/


alphabeta1234 wrote:
Why is the same formula (n+r-1)C(r-1) 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+4-1)C(4-1)?


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.
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Re: In how many ways can 5 different candiesbe distributed among   [#permalink] 05 Sep 2013, 20:37
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