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How many ways are there of placing 6 marbles in 4 bowls, if any number

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How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

source:gogmat
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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

whats wrong with D?

source:gogmat


Each marble has 4 options, so there are total of 4*4*4*4*4*4=4^6 ways.

Answer: C.
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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 12 Aug 2012, 12:28
why cant i use permutation?

have 4 slots: - - - - and for each slot there are 6 marbles, hence 6^4

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 12 Aug 2012, 15:56
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Galiya wrote:
why cant i use permutation?

have 4 slots: - - - - and for each slot there are 6 marbles, hence 6^4


Because there are more than "4 slots". By doing the slot method and saying that there are 4 bowls, hence 4 slots, you are saying that each bowl can only hold 1 marble. Problem is that a bowl can hold 1,2,3,4,5, or all 6 marbles. So instead of calculating how many options there are for the bowls to hold, it's simpler to find how many options each marble has.

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 12 Aug 2012, 18:17
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Galiya wrote:
why cant i use permutation?

have 4 slots: - - - - and for each slot there are 6 marbles, hence 6^4


You are distributing marbles to slots .. not slots to marbles !

The reason you cant use permutations is that you are not permuting ! It is a pure distribution problem. The distribution problems have a different strategy than permutation and combinations.

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

whats wrong with D?

source:gogmat



Forget about permutations and combinations, I mean what the process is called.
Think how would you do it:
Take the first marble and think what you can do with it. Where can you place it? - you have 4 options, as there are 4 bowls
Take the second marble - 4 options again, you don't care about the previous one already placed
Third marble - still 4 bowls available, still have the freedom to chose any one
...
You are choosing a bowl for each marble.
This will give you \(4^6\) possibilities.


D would be the correct answer for example if we have 6 bowls and 4 marbles.
first marble to place - 6 choices
second marble - again 6 choices
...
This will give you \(6^4\) possibilities.
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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 12 Aug 2012, 23:09
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EvaJager wrote:
Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

whats wrong with D?

source:gogmat



Forget about permutations and combinations, I mean what the process is called.
Think how would you do it:
Take the first marble and think what you can do with it. Where can you place it? - you have 4 options, as there are 4 bowls
Take the second marble - 4 options again, you don't care about the previous one already placed
Third marble - still 4 bowls available, still have the freedom to chose any one
...
You are choosing a bowl for each marble.
This will give you \(4^6\) possibilities.


D would be the correct answer for example if we have 6 bowls and 4 marbles.
first marble to place - 6 choices
second marble - again 6 choices
...
This will give you \(6^4\) possibilities.


Lets have a different problem
Their are 4 alphabets Set A ( A,B,C,D) and 6 Numbers SET N ( 1,2,3,4,5,6)
Q.1 : In how many ways can 4 alphabets be assigned a number from 1 to 6 without any restrictions
Q.2 : In how many ways can 6 numbers be assigned an alphabet from A to D without any restriction
These two should clear your doubt
and to make these easy start adding restrictions to them like use all elements of set A, Set N, map uniquely , no element used twice, etc
And these two questions can be formed in any way to show all concepts of Permutations, combinations, distributions, De-distribution and even to bose einstien distribution.

and can you think of a condition that will make this mapping a function ?

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 15 Aug 2012, 13:23
Bunuel wrote:
Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

whats wrong with D?

source:gogmat


Each marble has 4 options, so there are total of 4*4*4*4*4*4=4^6 ways.

Answer: C.



Bunuel:

In another excercise (I cannot attach the link because this is my second post and the system is not allowing me) you explained this:

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 is (n+r -1)C(r-1).


Following this statement, taking in mind Persons = Bowls, I have to think that the answer to this question is 9C3 = 9! / 3!6! = 84. But it is incorrect according to your post. Could you please explain a little further?

Thanks a lot, José

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 15 Aug 2012, 22:59
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Josefeg wrote:
Bunuel wrote:
Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

whats wrong with D?

source:gogmat


Each marble has 4 options, so there are total of 4*4*4*4*4*4=4^6 ways.

Answer: C.



Bunuel:

In another excercise (I cannot attach the link because this is my second post and the system is not allowing me) you explained this:

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 is (n+r -1)C(r-1).


Following this statement, taking in mind Persons = Bowls, I have to think that the answer to this question is 9C3 = 9! / 3!6! = 84. But it is incorrect according to your post. Could you please explain a little further?

Thanks a lot, José


It wasn't stated explicitly, but we all assumed in our solutions, that all the marbles are distinct/different (think of different colors or numbered marbles). Then the above solutions are correct. The number of possibilities to place 6 distinct/different marbles in 4 bowls is \(4^6.\)

If the marbles are all identical, the bowls are distinct, then what is different between the distributions is the particular number of marbles in each bowl.
In this case, the above formula you mentioned should be used. For example, 6 identical marbles can be placed in 4 bowls in (6 + 4 - 1)C(4 - 1) = 9C3 = 84 ways. In the original question, since none of the listed answers is 84, the hidden assumption was that the marbles are non-identical, which I think it should have been stated explicitly.

For \(n\) identical marbles and \(r\) bowls, a way to justify the formula is as follows: think of the of the marbles placed in slots instead of bowls. The slots are aligned, created such that there are \(r-1\) dividing internal walls, something like this: [o|ooo|...| |o| ], where [ and ] represent the two outer walls of the slots. In the first slot there is one ball, in the second three balls,..., there is an empty slot, just one ball, and the last one is also an empty slot.
In each slot, we can place any number of marbles between 0 and r.
Imagine that we have \(n+r-1\) places, because we have \(n\) marbles and \(r-1\) dividing walls, and we just have to decide in this string of length \(n+r-1\) where to place the walls (or equivalently, where to place the marbles). This can be done in (n + r - 1)C(r - 1) different ways, or equivalently, (n + r - 1)Cn.
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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 16 Aug 2012, 08:56
EvaJager wrote:

It wasn't stated explicitly, but we all assumed in our solutions, that all the marbles are distinct/different (think of different colors or numbered marbles). Then the above solutions are correct. The number of possibilities to place 6 distinct/different marbles in 4 bowls is \(4^6.\)



Eva, thanks a lot for your explanation. I also think that it should have been stated.

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 13 Jul 2014, 07:20
Quote:
Lets have a different problem
Their are 4 alphabets Set A ( A,B,C,D) and 6 Numbers SET N ( 1,2,3,4,5,6)
Q.1 : In how many ways can 4 alphabets be assigned a number from 1 to 6 without any restrictions
Q.2 : In how many ways can 6 numbers be assigned an alphabet from A to D without any restriction
These two should clear your doubt
and to make these easy start adding restrictions to them like use all elements of set A, Set N, map uniquely , no element used twice, etc
And these two questions can be formed in any way to show all concepts of Permutations, combinations, distributions, De-distribution and even to bose einstien distribution.

and can you think of a condition that will make this mapping a function ?


Wow! Nice explanation EvaJagger, Bunuel and mandyrhtdm!
A lot of doubts cleared from one single thread! Kudos to all

@mandrhtdm just to confirm..the answers are Q1 \(4^6\) Q2 \(6^4\) right?

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 06 Jan 2018, 04:27
Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

whats wrong with D?

source:gogmat


at first glance it is very essential to notice that "any number of marbles can be placed in each bowl"
since "0" is a number, the statement actually dictates that any bowl can get no marbles, and any bowl can get more than 1 marbles, since there are 6 marbles in total. So far so good.

now, for each marble, there are 4 ways to be distributed, as there are 4 bowls

1st marble can be distributed in 4 ways
2nd marble can be distributed in 4 ways
3rd marble can be distributed in 4 ways
4th marble can be distributed in 4 ways
5th marble can be distributed in 4 ways
6th marble can be distributed in 4 ways

which implies

4 * 4 * 4 * 4 * 4 * 4

= 4 ^ 6 is the answer

thanks
:cool:

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number [#permalink]

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New post 06 Jan 2018, 17:11
Galiya wrote:
How many ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?

A. 6C4
B. 6P4
C. 4^6
D. 6^4
E. 6!

source:gogmat

------------
6 marbles to be placed in 4 bowls. There are no restriction on the number of the marbles that the bowl can have:
so each marble has 4 options/bowls to choose from.
thus in total 4^6 no.of ways

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Re: How many ways are there of placing 6 marbles in 4 bowls, if any number   [#permalink] 06 Jan 2018, 17:11
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