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

Question

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

EvaJager · Accepted Answer

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.

EvaJager · Answer

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:  , where   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.

Bunuel · Answer

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

Answer: C.

galiya · Answer

why cant i use permutation?

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

Injuin · Answer

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.

mandyrhtdm · Answer

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.

mandyrhtdm · Answer

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 ?

Josefeg · Answer

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

Josefeg · Answer

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

testcracker · Answer

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

luffy_ueki · Answer

------------
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

akadiyan · Answer

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

Each marble has 4 options and can be placed in any bowls, so the answer is 4*4*4*4*4*4 =  4^6

Ans:   C

RenB · Answer

Can 'any' number imply 'zero' as well? It appears to be a valid way of placing the marbles given that any number of marbles can be so placed.
Then each marble will have 5 options thus answer would be 5^6

Let me know if I am overthinking and if this is actually an ambiguity; also if this type of ambiguity can be seen on the exam.

Thanks!

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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

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