Combinatorics: a set of 30 from unlimited objects

1. Mar 2, 2017

Lord Anoobis

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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?

2. Mar 2, 2017

Dick

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.

3. Mar 2, 2017

Lord Anoobis

Two colours would lead to 31 possible combinations and 1770 the way I used.

4. Mar 2, 2017

Dick

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?

5. Mar 2, 2017

Lord Anoobis

Yes indeed, four groups making up a total of thirty which means we must have $30 \choose 4$, correct?

6. Mar 2, 2017

Dick

Closer. Think again about the two color case. How would you write that in combinatorial form?

7. Mar 2, 2017

Lord Anoobis

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.

8. Mar 2, 2017

Lord Anoobis

All I can think of is $31 \choose 1$ but I don't see the reasoning behind it.

9. Mar 2, 2017

Dick

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?

10. Mar 2, 2017

Lord Anoobis

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. Mar 2, 2017

PeroK

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. Mar 2, 2017

Lord Anoobis

I think a complete restart is in order. I'll get back to this in a bit, work beckons unfortunately.

13. Mar 2, 2017

Dick

$30 \choose 2$ would be taking two balls out of 30, leaving three groups totalling 28 balls.

14. Mar 2, 2017

haruspex

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.

15. Mar 2, 2017

Lord Anoobis

Will do. It's close to midnight in this part of the world so that's a mission for the morrow.