Can you define a numbering system for these combinations?

unique numbers

= 1 + C(n1 - 1 , 1) + C(n2 - 1 , 2) + C(n3 - 1 , 3) + C(n4 - 1 , 4) + C(n5 - 1 , 5) + C(n6 - 1 , 6) + C(n7 - 1 , 7) + C(n8 - 1 , 8) + C(n9 - 1 , 9)

number 1 is 111111111000000000000000000 and number 4686825 is 000000000000000000111111111.

n1 is at the leftmost (as n9 is at the rightmost) and for any C(n,r) with n < r, it is 0.

the formula for x balls within y slots is

= 1 + ΣC(n_i - 1 , i) , with i = [1,x] and n_i = [1,y]

Let the permutation be

a[27] a[26] a[25] . . . a[3] a[2] a[1],

where a[i]=0 or a[i]=1.

Let the number of '1' be 9.

Let k be a non-negative integer, such that:

1) 0<=k<=9

2) for i=1 to k, all a[i]=1

3) a[k+1]=0

Then the unique identifying number N of each permutation will be:

N=1+∑ [i=k+1 to 27] a[i]*(i-1)!/{u[i]!*(i-1-u[i])!}

where u[i] is the number of '1' from a[1] to a[i]

It's easy to write a computer program based on the above logic.

If I have time, I'll try to write it.

Edit:

Examples:

a) 000000000000000011101110111

k=3

N=1+10!/(9!1!)+9!/(8!1!)+8!/(7!1!)+

6!/6!+5!/5!+4!/4!=31

b) 100010001000100010001001110

k=0

N=1+26!/(9!17!)+22!/(8!14!)+

18!/(7!11!)+14!/(6!8!)+10!/(5!5!)+

6!/(4!2!)+3!/3!+2!/2!+1!/1!

c) 111111111000000000000000000

k=0

N=26!/(9!17!)+25!/(8!17!)+24!/(7!17!)+

23!/(6!17!)+22!/(5!17!)+21!/(4!17!)+

20!/(3!17!)+19!/(2!17!)+18!/(1!17!)+1=

4686825

Try this:

Let the number of slots be n, in this instance, 27, and the number of balls be k, in this instance, 9. Let the position p in your combinatorial "number" be zero for the least significant digit and k−1 for the most significant. The least significant number will represent the position of the highest ball, the next least significant be the next highest, and so on. Each digit in this representation will be to the base d(p−1)+1, where d(p) is the digit in position p. d(0) is in base n−k. To generate the number:

1. Set your number accumulator be zero.

2. For p = 0 to k−1 do:

3. If m is the position of the (p+1)'th highest ball, set the digit at position p to m−k+p. If this digit is zero, you may optionally stop early and return the contents of the accumulator because all less significant digits will also be zero.

4. Next p.

5. Return the contents of the accumulator.

The p'th digit of this representation is in base d(p−1)+1, where d(p) is the digit in position p. Position zero in is in base n−k. To decode this number:

1. Set the accumulator to zero.

2. For p = k−1 to 1 do:

3. Set the base counter b to d(p−1)+1

4. Multiply the accumulator by b and add d(p).

5. Next p.

6. Set the base counter b to n−k.

7. Multiply the accumulator by b and add d(0)

8. Return the contents of the accumulator.

i had deleted all.

---

now the question are clear, thanks.

i'll present my algorithm.

if you'll understand it then you realized that it is based on the identity sum_{i=0}^{t} {(n+i over n)} = (n+t+1 over n+1), for any positive integers n and t fixed.

you have (27 over 9) combinations. it's ok.

consider those number in base 2, so they are composed of nine numbers 1 and eighteen numbers 0, randomly.

now we can assign (by bjiection) each of those number to a number x in the intervall [1, (27 over 9)].

we need to proceed like that to construct x:

- start with 1.

CASE A

if the first digit in base 2 is 1 then you dont need to add anything;

- if the second digit is 0 then you must add the number of possibe numbers which starts by (1,1) which are (25 over 7);

- if the second digit is 1 then you dont need to add anything.

CASE B

otherwise the first digit is 0, so you must add the number of possible numbers which starts by 1 (so it is (26 over 9)):

- of the second digit is 1 then you dont add anything;

- if the second digit is 0 then you must add the number of possibl numbers which starts by (0,1) which are (25 over 8)..

and so on..

it is important to understand the meaning of the binomial coefficient...we can prove that this algorith generate effectively a bjiection: in fact we can start from the image, e.g. the number 10000.

so we need to proceed like that:

(10000-1)<(26 over 9) so the first digit is 1(in base 2).

(10000-1)<(25 over 8) so the second digit is 1.

(10000-1)<(24 over 7) so the third digit is 1.

(10000-1)<(23 over 6) so it is again 1.

(10000-1)<(22 over 5) so it is again 1.

(10000-1)>(21 over 4) so the sixth digit is 0. so we need to delete the (21 over 4) itself.

(10000-1-(21over 4)) =4014<(20 over 4) so we have a 1.

4014 > (19 over 3) so we have a 0 and we must consider the number 4014-(19 over 3)= 3045.

and so on, the process is unique, so it is unique also te number in base 2 and the associated combination of 9 balls in27 slots.

cheers

paolo.

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