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

  • #1

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)
 

Answers and Replies

  • #2
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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?
 
  • #3
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.
 
  • #4
Nathanael
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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?
 
  • #6
haruspex
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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.
 
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  • #7
haruspex
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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.
 
  • #8
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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 cant 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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