# My math question

Discussion in 'Scientific Statistics Math' started by PortlandDream, Dec 30, 2011.

1. ### PortlandDreamGuest

Hello,

I was wondering if there might be a formula to solve this problem I
devised.

A = number of people
B = number of apples
C = total possible variations

Where you can assign to all values of A all different values of B (so
that the sum always ends up being B) to get the total variations.

I was kinda dreaming about an apple a day and thought if I could
spread it out, or if someone hogged it all plus all variations. Yeah I
am a math nerd.

I thought I have seen a formula for this, but I cannot recall it for
the life of me.

I have made some tables for this but it is clear the system will
spiral out of hand without a formula and I cannot derive one so far
with the information at hand.

Any help would be awesome!

PortlandDream, Dec 30, 2011

2. ### Rich UlrichGuest

Obsessing about combinations has never been one of my
hobbies, but here are a few comments.

Sometimes the solutions to combinatorial problems seem to
"spiral out of hand" even when you have a formula -- if that
means "a lot of tedious detail" when you are not talking about one
of the extreme cases.

For a better chance for help with this particular problem --
You need to be more specific about what you want to count, the
detailed layout of the problem. I *think* that you want to
distribute B indistinguishable apples among A indistinguishable
people; but otherwise, it is a different problem. Are you talking
about small numbers, or *very* small numbers, or something else?
- If person #1 has all the apples, is that same solution as if
person #2, or #3, or ..., has all the apples?

Tedious detail:
(5,...), (4,1,...), (3,1,1,...), (3,2, ...), (2, 1,1,1, ...), (2,2,1),
(1,1,1,1,1).

If the people have labels, there more answer sets, and
you are probably interested in a tiny subset of possible
questions.

Hope this helps.

Rich Ulrich, Jan 1, 2012

3. ### PortlandDreamGuest

Umm, the hardest version with each person named.

So if three apples, three people, for example, we would see 9 results

(3 . .) (. 3 .) (. . 3) (2 1 .) (2 . 1) (1 2 .) (. 2 1) (1 . 2) (1 1
1)

If I just could remember the name of the method I could sleep in
peace

PortlandDream, Jan 2, 2012
4. ### Rich UlrichGuest

Or, it may be considered simpler to list those answers
because you don't have to find the "similar" cases
Combinatorials.
Combinatorics formulas.

Rich Ulrich, Jan 2, 2012
5. ### Jussi PiitulainenGuest

Borrow two oranges and prepare a row of three plus two positions. In
each position, put either an apple or an orange. Every way of doing
this creates three groups of apples, separated by the oranges.

The number of the arrangements is the number of ways to choose the
positions for the oranges.
No name comes to mind.

Jussi Piitulainen, Jan 2, 2012