How many ways can 12 balls be arranged into 4 different rows

[kukumaluboy]

Homework Statement

In how many ways can 12 balls be arranged into 4 different rows with each row having
at least one ball
(a) if the balls are identical?
(b) if there are 6 identical red balls and 6 identical blue balls?

Homework Equations

The Attempt at a Solution

a)
Put 4 balls in each row.
Remaining 8 balls left.
Label each row A, B, C, D
A + B + C + D = 8 balls
Ans: 11C3 = 165 (Correct)

b) (Unsure how to do)

 

[jedishrfu]

You have to look at your first answer and divide out but the number of arrangements.

As an example, if I have abcd and I want to know all the words I can make the answer is 4!

But if the d letter is an a then I have to divide my count by 2 a1 b c a2 is the same as a2 b c a1 right?

So in your what must you do?

 

[member 587159]

known data

In how many ways can 12 balls be arranged into 4 different rows with each row having
at least one ball
(a) if the balls are identical?
(b) if there are 6 identical red balls and 6 identical blue balls?

You have to look at your first answer and divide out but the number of arrangements.

As an example, if I have abcd and I want to know all the words I can make the answer is 4!

But if the d letter is an a then I have to divide my count by 2 a1 b c a2 is the same as a2 b c a1 right?

So in your what must you do?

Are you sure about this? Divide out? Intuitively, I would say there will be more options because not all balls are identical. I'm not sure though so sorry if I'm mistaken.

 

[Nathanael]

May someone please explain to me why the answer to (a) is 11choose3? I agree with it, but I got the answer by way of a nested sum, namely $\sum\limits_{n=1}^{9}\sum\limits_{m=1}^{n}m$ ... I'm curious, what perspective brings out 11choose3?

 

[member 587159]

https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition

May someone please explain to me why the answer to (a) is 11choose3? I agree with it, but I got the answer by way of a nested sum, namely $\sum\limits_{n=1}^{9}\sum\limits_{m=1}^{n}m$ ... I'm curious, what perspective brings out 11choose3?

 

[haruspex]

May someone please explain to me why the answer to (a) is 11choose3? I agree with it, but I got the answer by way of a nested sum, namely $\sum\limits_{n=1}^{9}\sum\limits_{m=1}^{n}m$ ... I'm curious, what perspective brings out 11choose3?

@Math_QED's link is certainly applicable, but here's a more intuitive approach.
We have n identical objects to place in k distinct buckets. We can place the objects in a line and represent the buckets by drawing boundaries between the objects at k-1 places. We can then think of this arrangement as n+k-1 things, of which some n are the objects. Each possible distribution of the n objects into the k buckets corresponds 1-1 with a decision about which n of the n+k-1 things are objects. Thus the number of distributions is the number of ways of choosing n things from n+k-1 things.

 

[haruspex]

Are you sure about this? Divide out? Intuitively, I would say there will be more options because not all balls are identical. I'm not sure though so sorry if I'm mistaken.

Agreed.

 

[Biker]

The way I solved it is very similar.

Assume that you place them in one line and you have 3 brackets (To divide it into 4 groups) So you pretty much have 13 places to place the brackets... However, I can't place them at both ends because it has to have 1 for each group. So 13-2 = 11
Now choose 3 places from 11
By
11c3