Symmetry - II

I just love solving things by symmetry. I feel as the variables increase in any problems, symmetry can play bigger and bigger role. Another CAT type question:
The average value of Mod (a - b) + Mod (c - d) + Mod (e - f) for all possible permutations a, b, c , d, e, f of 1, 3, 5, 7, 9, 11 is :
My first reaction when I saw this problem, I will not attempt this unless I have no other question left. But once again I thought so many variables - must be some easy way out. I feel these problems can be solved in steps making them simpler with each step. I mean what is the use of Mod here - when values of a or b and c or d and e or f can be interchanged. As in once we choose the pairs (a,b), (c,d) and (e,f) such that values of (a and b), (c and d) and (e and f) can be interchanged - we can form 2*2*2 combination of the expression with interchanging values of and b or c and d or e and f. Also, because of Mod function outcome of the expression for all 8 ways would be the same. Out of these 8 case there surely will be one and only one case where a>b, c>d and e>f. Hence, if we just choose this one case and ignore 7 other cases as they do not add or subtract anything to symmetry as outcome of all the 8 are same and when we average it, it is equivalent to multiply for one particular a - b + c - d + e - f by 8 and then dividing by 8. Hence the expression can be reduced as (a-b) + (c-d) + (e-f) where a>b, c>d and e>f.
Hence, now we have to find Average (a-b + c-d + e-f) which can further be simplified to Average (a-b) + Average (c-d) and Average (e-f).
Now the beautiful symmetry: Average (a-b) must be equal to Average (c-d) must be equal to Average (e-f)
So the expression reduces to 3* Average (a-b)
Now the problem has just reduced to plug the numbers and get the average.
If a=11, b can be 1,3,5,7 and 9 and (a-b) = (10,8,6,4 and 2)
If a=9, b can be 1,3,5 and 7 and (a-b) = (8,6,4 and 2)
If a=7, b can be 1,3 and 5 and (a-b) = (6,4 and 2)
If a=5, b can be 1 and 3 and (a-b) = (4 and 2)
If a=3, b can be 1 and (a-b) = 2
Hence average = 3*((10,8,6,4 and 2)+(8,6,4 and 2)+(6,4 and 2)+(4 and 2)+2))/(5+4+3+2+1)
= 3*70/15 = 14 and ofcourse which is the correct answer!
This problem should take two minutes but solvingthese kind of problems do help in making the concepts clear in your mind. This problem appeared difficult in the beginning, but I always believe that if its unsolvable in two minutes, it will not be in the CAT paper. So there is an easy way out always - practice and deriving solutions on your own with your own methods and technique will surely help in finding that way.
Just to post the alternate solution on some CAT related website which I found was a bit complex : By symmetry the average of a - b is independent of choices in which they are chosen (provided they are unequal). Suppose a = k. Then the average of k - b is (k-1 + k-3 + ... + 2 + 2 + 4 + ... + 11-k)/5 = 2*(k^2 -7k + 21)/5. So average of a - b is (1/6)*2*sigma (k^2 -7k + 21)/5. Hence required average is 2*sigma(k^2 -7k + 21)/10 = 2*(91 - 147 + 126)/10 = 14.