# Statistics Problem

1. Jan 27, 2008

### kuahji

Statistics Problem :(

An art collector, who owns 10 paintings by famous artists, is preparing her will. In how many different ways can she leave these paintings to her three heirs?

Reasoning this out, she can leave 10 to the first heir & 0 to the others, 9 two the first heir & 1 to either of the others, etc. But, without creating a tree diagram which would be timely, how can I represent this? The back of the book gives an answer of 59,049 ways, but I'm just lost regarding how to set everything up.

2. Jan 27, 2008

### Dick

Each painting can be left to one of three heirs. So there are three choices for each painting and there are 10 paintings. Do you get my drift?

3. Jan 27, 2008

### kuahji

Ok, I finally punched the right numbers into the calculator... don't think I'll ever truly get statistics.

Basically what I did was
3nCr1=3
then 3^10 = 59049
Now that I came up with that, it makes some sense. Thanks for the help.

4. Jan 27, 2008

### kuahji

This is what drives me batty, you use the same logic on a similar problem & it doesn't work.
If a bakery has 12 apple pies left at the end of a given day, in how many different ways can it distribute these pies among six food banks?

What exactly is different about this problem from the last? I'm not asking for the answer or anything like that (which happens to be 6,188). Rather just what is different in this one than the last one?

I tried 6nCr1=6
6^12=2176782336. Clearly its not correct reasoning.

5. Jan 27, 2008

### Dick

The difference is that the paintings are distinguishable (they are different). The pies are not. Work it out based on that. That's rather more like your first analysis.

6. Jan 28, 2008

### kuahji

Thanks for pointing that out, it was a rather difficult problem. Finally after searching the internet for hours I found an equation that works.

(n+r-1)C(r-1)
Which gives 17C5 = 6188.

Though sadly I have no idea why it works, just that it does.

7. Jan 28, 2008

### Dick

If it makes you feel any better I had to scratch my head over that also, guess I'm out of practice. Picture 17 objects lined up in a row. Pick 5 of them. Call the objects that weren't picked 'pies'. Call the objects that were picked 'separators'. If you scratch you head like I did, after while it will dawn on you that the number of choices of separators is the same as the number of ways to partition the 12 pies.

Last edited: Jan 28, 2008