Combinatorics / Permutation problem

[FlawedChip]

I have a certain number of items, 26 (A-Z). I would like to find the number of possible combinations by splitting up 100% (1-100) among the 26 items.

Here are a couple of outcomes that would exist for example:

Item Percentage
A: 100
B-Z: 0

Item Percentage
A: 99
B: 1
C-Z: 0

Item Percentage
A: 98
B: 1
C: 1
D-Z: 0

Item Percentage
A: 0
B: 15
C: 30
D-L: 0
M: 20
N: 25
O-Y: 0
Z: 10

Is there a formula to use to count the number of outcomes? I think it might be possible using nCr or nPr?

[TimM]

Looks simple to me. Consider each letter as a bin, and each percentage point as an identical marble to be placed in one of the bins.

[jay_shark]

This is easy .

You're just looking for the number of nonnegative integer valued solutions to the following :

a1+a2+a3+...+a26 = 100

The number of solutions is : (100+ 26-1)C(26-1)=~1.30X10^26

[jay_shark]

Here is the formula that I will derive .

There are (n+r-1)C (r-1) nonnegative integer valued solutions a1,a2,a3,...ar satisfying

a1+a2+a3+...ar = n

Take n+r O's : 000...000 and note that there are n+r-1 spaces between consecutive 0's .
Now divide the O's using dashes and choose r-1 dashes .

ie n=4 r=2 0|00000 implies a1=0 and a2=4.
00|0000 implies a1=1 and a2=3 .

n=5 r=3 00||000000 implies a1=1 a2=0 a3=4

[jay_shark]

Here is a polished solution .

Take n 0's and r-1 dashes .

n=5 r=3 00000|| This tells us that a1=5 a2=0 a3=0

n=7 r=4 00000||0|0 This tells us that a1=5 ,a2=0 and a3=1 and a4=1 .