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<SPOILER>  (maybe...)
























































Is it 351?  I figured it this way:

Call the number of apples given to person 1 N, person 2 X, and person 3 Y.
Now if N = 25, then X and Y must both equal 0. (1 solution)
If N = 24, then {X=1,Y=0} and {X=0, Y=1} are valid. (2 solutions)
If N = 23, then {X=1, Y=1}, {X=0, Y=2}, {X=2, Y=0} are valid. (3 solutions)
If N = 22, then there are 4 solutions.
If N = 21, then there are 5 solutions.
...
If N = 0, then there are 26 solutions (run X from 0 to 25 and Y is
forced to a value).  
So the total number is 1+2+3+4+...+26 or 351.
I'm rather math-impaired, so I would like to know 1) is this right? 2)
what is a
more "scientific" way of finding the answer.

Brian /-|-\

There is a formula (Ramanujan's Formula, I believe), that lets you
figure out the number of ways you can divide a number -- ie. the number
of different ways you can make a series of numbers that adds up to the
number in question.  I don't know the formula offhand :-) and I don't
know how useful it would be for this particular question, however.

--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have realized it would be bad vodka when all the label said
was 'Russian Vodka'."

bcash@crchh410.BNR.CA (Brian Cash) writes:

><SPOILER>  (maybe...)

>Is it 351?  I figured it this way:

I think so, and I worked it the same way. I'm now trying to find a general
solution for dividing Q apples among P people - I'll post if I find one.

>Call the number of apples given to person 1 N, person 2 X, and person 3 Y.
>Now if N = 25, then X and Y must both equal 0. (1 solution)
>If N = 24, then {X=1,Y=0} and {X=0, Y=1} are valid. (2 solutions)
>If N = 23, then {X=1, Y=1}, {X=0, Y=2}, {X=2, Y=0} are valid. (3 solutions)
>If N = 22, then there are 4 solutions.
>If N = 21, then there are 5 solutions.
>...
>If N = 0, then there are 26 solutions (run X from 0 to 25 and Y is
>forced to a value).  
>So the total number is 1+2+3+4+...+26 or 351.

Christopher Hicks
--
 ==============================================================================
 Christopher Hicks    |    If it doesn't fit...
 cmh@uk.ac.cam.eng    |                        ...you need a bigger hammer.
 ==============================================================================

>From: padmanab@ecs.umass.edu
>Here is a SIMPLE problem I came across long ago. 
>I am posting it because I think it has a wonderful 
>solution. 
>
>Suppose there are 25 identical apples. How many 
>ways can you divide them among three people? Also, 
>all the apples must be divided among them. 

In article <C1E5Du.D3B@news.rich.bnr.ca> bcash@crchh410.BNR.CA (Brian Cash) writes:
>Is it 351?  I figured it this way:
> (etc...)
>I'm rather math-impaired, so I would like to know 1) is this right? 2)
>what is a more "scientific" way of finding the answer.

The really nice way that I think the poser refers to is to think
of picking two distinct numbers from 0 to 26 inclusive.
Call the smaller x and the larger y.
Then the first person gets x apples, the second y-x-1 apples,
and the third 26-y.  Note the most anyone can get is 25, the least 0,
and the total is also 25.
The number of ways of doing each of these things is the same.
                    27
In this case, it is   C   or 27x26/2 or 351.
                        2
This also generalizes easily, e.g. dividing 30 apples among 5 can be
     34
done   C    ways.
         4
Incidentally, I first came across this problem when studying quantum
mechanics, which was the last place I expected to see combinatorics.
___________________________________________________________________
Tim Adam, B.Sc./B.E.(Elec) student, Melbourne University, Australia
Ph: (03) 807 3279   E-mail: tmadam@ee.mu.oz.au