MHB Placing Balls in Numbered and Unnumbered Boxes: Infinite Possibilities

[evinda]

Hey! (Giggle)

I am given this exercise:
If we have a pile of infinite balls and $n$ numbered boxes with a capacity of $n$ balls each one,with how many ways can we place some balls in the boxes?Answer the same question,if the boxes are not numbered.

Could you give me a hint what to do?? (Blush)

 

[evinda]

Is it maybe $n^n$ ?? Or am I wrong?

 

[I like Serena]

Hey! (Giggle)

I am given this exercise:
If we have a pile of infinite balls and $n$ numbered boxes with a capacity of $n$ balls each one,with how many ways can we place some balls in the boxes?Answer the same question,if the boxes are not numbered.

Could you give me a hint what to do?? (Blush)

Is it maybe $n^n$ ?? Or am I wrong?

Hi! (Smirk)

How did you get $n^n$?

 

[evinda]

Hi! (Smirk)

How did you get $n^n$?

I thought it,because of the fact that there are $n$ numbered boxes,and at each one we can put $n$ balls..Is it wrong?? (Thinking)

 

[I like Serena]

I thought it,because of the fact that there are $n$ numbered boxes,and at each one we can put $n$ balls..Is it wrong?? (Thinking)

Let's start with 1 box that can contain $n=3$ balls.
What are the possibilities to fill it? (Wondering)

 

[evinda]

Let's start with 1 box that can contain $n=3$ balls.
What are the possibilities to fill it? (Wondering)

With infinite balls,or not??

 

[I like Serena]

With infinite balls,or not??

Yes. (Wasntme)
And I guess we'll have to assume those infinite balls are identical, or the problem becomes a bit non-sensical otherwise.

 

[evinda]

Yes. (Wasntme)
And I guess we'll have to assume those infinite balls are identical, or the problem becomes a bit non-sensical otherwise.

So,are there infinite ways?? Or is there an other formula,that expresses it?? (Thinking)(Thinking)

 

[I like Serena]

So,are there infinite ways?? Or is there an other formula,that expresses it?? (Thinking)(Thinking)

If the infinite balls are unique, there would indeed be infinite ways.
That is why I am assuming that they can not be distinguished from each other. (Wink)

So 1 box with capacity 3 could for instance contains 3 balls.
What are the other possibilities?
How many are those?

 

[evinda]

If the infinite balls are unique, there would indeed be infinite ways.
That is why I am assuming that they can not be distinguished from each other. (Wink)

So 1 box with capacity 3 could for instance contains 3 balls.
What are the other possibilities?
How many are those?

So,is it also possible that 1 box contains also $2$, $1$ or $0$ balls?
But how can I find then the number of ways we can place some balls in the boxes?? (Thinking)

 

[I like Serena]

So,is it also possible that 1 box contains also $2$, $1$ or $0$ balls?
But how can I find then the number of ways we can place some balls in the boxes?? (Thinking)

Now I am going to assume the box does not have specific places for the balls, but that it just contain a number of balls.

So for 1 box with capacity $n=3$ we have $4$ ways to fill it... (Thinking)

 

[evinda]

Now I am going to assume the box does not have specific places for the balls, but that the just contain a number of balls.

So for 1 box with capacity $n=3$ we have $4$ ways to fill it... (Thinking)

So,one box with capacity $n$ has $n+1$ ways to be filled.So,in general, $n$ boxes have $(n+1)^n$ ways to be filled,right??

 

[I like Serena]

So,one box with capacity $n$ has $n+1$ ways to be filled.So,in general, $n$ boxes have $(n+1)^n$ ways to be filled,right??

Right! (Cool)

 

[evinda]

Right! (Cool)

Great!Thank you very much! (Clapping)

 

[DavidCampen]

Hey! (Giggle)

I am given this exercise:
If we have a pile of infinite balls and $n$ numbered boxes with a capacity of $n$ balls each one,with how many ways can we place some balls in the boxes?Answer the same question,if the boxes are not numbered.

For the second question, I interpret this as being equivalent to asking: "how many unique sets of n objects can be formed by selecting zero or more of each of n distinct symbols".

So, for n=2 we have R(2)=3: aa, ab, bb

and in general, for n>3, we have:
\[R(n)=\sum_{m=1}^{n}mS(n-m+1)=R(n-1)+\sum_{m=1}^{n}S(m)\]

where \[S(n)=\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\]