The following sum is constituted for each permutation p
1, p
2, p
3,..., p
9, p
A, p
B, p
10 of the duodecimal (base 12) integers 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, 10:
|p1-p2|+|p3-p4|+|p5-p6|+|p7-p8|+|p9-pA|+|pB-p10|
Determine the average value of all such sums.
Every one of the integers 1 - 12 (decimal) will be equally represented as each of p1 through p12 (decimal subscripts). So we need only the average of the 66 possible differences of pairs (pair members are never equal, so unequal likelihood of disparate vs same integers is not a consideration) and multiply by 6:
--pair-- difference
1 2 1
1 3 2
1 4 3
1 5 4
1 6 5
1 7 6
1 8 7
1 9 8
1 10 9
1 11 10
1 12 11
2 3 1
2 4 2
2 5 3
2 6 4
2 7 5
2 8 6
2 9 7
2 10 8
2 11 9
2 12 10
3 4 1
3 5 2
3 6 3
3 7 4
3 8 5
3 9 6
3 10 7
3 11 8
3 12 9
4 5 1
4 6 2
4 7 3
4 8 4
4 9 5
4 10 6
4 11 7
4 12 8
5 6 1
5 7 2
5 8 3
5 9 4
5 10 5
5 11 6
5 12 7
6 7 1
6 8 2
6 9 3
6 10 4
6 11 5
6 12 6
7 8 1
7 9 2
7 10 3
7 11 4
7 12 5
8 9 1
8 10 2
8 11 3
8 12 4
9 10 1
9 11 2
9 12 3
10 11 1
10 12 2
11 12 1
---
286
The 66 differences of pairs add up to 286 for an average of 13/3. The sum of six such numbers averages 26 (that's 22 in base-12).
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Posted by Charlie
on 2022-12-19 07:41:38 |
solution
