Combinatorics: a set of 30 from unlimited objects

  • Thread starter Thread starter Lord Anoobis
  • Start date Start date
  • Tags Tags
    Combinatorics Set
Click For Summary

Homework Help Overview

The problem involves determining the number of ways to choose 30 balls from an unlimited supply of blue, red, green, and white balls, allowing for any combination of colors. The context is combinatorics.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various interpretations of the problem, including viewing it as a partitioning issue and considering simpler cases with fewer colors. There are attempts to relate the problem to binomial coefficients and combinatorial reasoning.

Discussion Status

The discussion is ongoing, with participants exploring different methods and questioning their reasoning. Some have suggested starting over to rethink the problem, while others have pointed out potential misunderstandings in the application of combinatorial principles.

Contextual Notes

There is mention of the need to transform the problem into a new form for accurate counting, and participants are encouraged to consider simple examples to identify patterns. The discussion reflects a lack of consensus on the correct approach and highlights the complexity of the problem.

Lord Anoobis
Messages
131
Reaction score
22

Homework Statement



In how many ways can you choose 30 balls from an unlimited number of blue, red, green and white balls if you can choose any number of the different coloured balls?

Homework Equations

The Attempt at a Solution


What I did is view the problem as choosing from a set of 30 of each of blue, red, green and white balls which gives us 120, seeing that we must have ##b+r+g+w = 30##. Then the number of distinct sets is ##120 \choose 30##. Is this line of thinking correct?
 
Physics news on Phys.org
Lord Anoobis said:

Homework Statement



In how many ways can you choose 30 balls from an unlimited number of blue, red, green and white balls if you can choose any number of the different coloured balls?

Homework Equations

The Attempt at a Solution


What I did is view the problem as choosing from a set of 30 of each of blue, red, green and white balls which gives us 120, seeing that we must have ##b+r+g+w = 30##. Then the number of distinct sets is ##120 \choose 30##. Is this line of thinking correct?

No, not correct. Suppose there were only 2 colors. How many different ways to choose? You should be able to figure it out easily, just by thinking about it. Now try using your method and compare the numbers.
 
Dick said:
No, not correct. Suppose there were only 2 colors. How many different ways to choose? You should be able to figure it out easily, just by thinking about it. Now try using your method and compare the numbers.
Two colours would lead to 31 possible combinations and 1770 the way I used.
 
Lord Anoobis said:
Two colours would lead to 31 possible combinations and 1770 the way I used.

Ok, so can you think of a way to get the right answer? This is sort of like partitioning the balls into groups, isn't it? Does that ring a bell?
 
Yes indeed, four groups making up a total of thirty which means we must have ##30 \choose 4##, correct?
 
Lord Anoobis said:
Yes indeed, four groups making up a total of thirty which means we must have ##30 \choose 4##, correct?

Closer. Think again about the two color case. How would you write that in combinatorial form?
 
I can see that ##30 \choose 2## does not lead us to 31 if the previous reasoning is followed but I'm not seeing the why of it.
 
Dick said:
Closer. Think again about the two color case. How would you write that in combinatorial form?
All I can think of is ##31 \choose 1## but I don't see the reasoning behind it.
 
Lord Anoobis said:
All I can think of is ##31 \choose 1## but I don't see the reasoning behind it.

Imagine a line of 31 balls. Take one ball away. That leaves you with 30 balls split into two groups, those to the left of the missing ball and those to the right. The number of ways to split 30 balls into two groups is the answer to the problem with two colors. Clear enough?
 
  • Like
Likes   Reactions: PeroK
  • #10
Dick said:
Imagine a line of 31 balls. Take one ball away. That leaves you with 30 balls split into two groups, those to the left of the missing ball and those to the right. The number of ways to split 30 balls into two groups is the answer to the problem with two colors. Clear enough?
Hang on. ##30 \choose 2## would apply if we where choosing from 30 different objects. In the case of two colours we are only dealing with 31 distinct sets.
 
  • #11
Lord Anoobis said:
Hang on. ##30 \choose 2## would apply if we where choosing from 30 different objects. In the case of two colours we are only dealing with 31 distinct sets.

You need to think more. Your approach at the moment seems to be to take the last two numbers that came up and form a binomial coefficient with them! In this case, you have to really think about transforming the problem into a new problem. Only then can you start counting.
 
  • #12
PeroK said:
You need to think more. Your approach at the moment seems to be to take the last two numbers that came up and form a binomial coefficient with them! In this case, you have to really think about transforming the problem into a new problem. Only then can you start counting.
I think a complete restart is in order. I'll get back to this in a bit, work beckons unfortunately.
 
  • #13
Lord Anoobis said:
Hang on. ##30 \choose 2## would apply if we where choosing from 30 different objects. In the case of two colours we are only dealing with 31 distinct sets.

##30 \choose 2## would be taking two balls out of 30, leaving three groups totalling 28 balls.
 
  • Like
Likes   Reactions: Greg Bernhardt
  • #14
Lord Anoobis said:
I think a complete restart is in order. I'll get back to this in a bit, work beckons unfortunately.
The transformation PeroK refers to is very elegant, but subtle, and some don't get it even when it is explained to them.
I suggest starting with several very simple examples to see if you can see a pattern.
 
  • Like
Likes   Reactions: PeroK
  • #15
haruspex said:
The transformation PeroK refers to is very elegant, but subtle, and some don't get it even when it is explained to them.
I suggest starting with several very simple examples to see if you can see a pattern.
Will do. It's close to midnight in this part of the world so that's a mission for the morrow.
 

Similar threads

Choose a ball at random from a randomly selected box - alternative way
  • · Replies 10 ·
Replies
10
Views
3K
Probability Problem: Choosing White Balls from a Mixed Bag
  • · Replies 10 ·
Replies
10
Views
3K
Simple probability question on combinatorics
  • · Replies 1 ·
Replies
1
Views
4K
Minimum numbers of balls to be drawn
  • · Replies 2 ·
Replies
2
Views
3K
High School Relative And Absolute Probability (Probability Of Picking A Red Ball)
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Billard balls on table
  • · Replies 10 ·
Replies
10
Views
3K
Probability: Draw w/o replacement, 3 different kinds of balls, don't understand
  • · Replies 3 ·
Replies
3
Views
3K
High School Probability - two possible points of view?
  • · Replies 7 ·
Replies
7
Views
2K
How many sets of 6 can be made from a set of 12
  • · Replies 18 ·
Replies
18
Views
3K
Arranging Identical Chips in a Circle: Combinatorics Question Explained
  • · Replies 4 ·
Replies
4
Views
4K