My only way is to try to list them; this is nasty!
Case 1: the subset contains the number 6
A) You choose five numbers lower than 6 (1 way)
and five numbers larger than 6 (1 way) for a total of 1 way
B) You choose four numbers lower than 6 (5 ways)
and four numbers larger than 6 (5 ways) for a total of 5 x 5 = 25 ways
C) You choose three numbers lower than 6 (10 ways)
and three numbers larger than 6 (10 ways) for a total of 10 x 10 = 100 ways
D) You choose two numbers lower than 6 (10 ways)
and to numbers larger than 6 (10 ways) for a total of 10 x 10 = 100 ways
E) You choose one number lower than 6 (5 ways)
and one number larger than 6 (5 ways) for a total of 5 x 5 = 26 ways
F) You choose no number lower than 6 (1 way)
and no number larger than 6 (1 way) for a total of 1 x 1 = 1 way
Now, for the really nasty part, you don't choose the number 6 -- when you do this,
the largest number below 6 and the smallest number above 6 must have a mean
of 6.
A) You choose five numbers lower than 6 (1 way)
and five numbers larger than 6 (1 way) for a total of 1 way
B) You choose four numbers lower than 6 and four numbers larger than 6 (10 ways)
C) You choose three numbers lower than 6 and three numbers larger than 6 (46 ways)
D) You choose two numbers lower than 6 and two numbers larger than 6 (30 ways)
E) You choose one number lower than 6 and one number larger than 6 (5 ways)
Hopefully, I counted correctly -- if not, maybe someone can correct me ...
