# 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?

## 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)

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?

kukumaluboy said:
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?
jedishrfu said:
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.

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?

• Nathanael
Nathanael said:
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.

• Biker and Nathanael
Math_QED said:
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.

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

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