This seemed like a formidable problem,
. . but with a little thinking . . .
Nearly all six-digit numbers have an equivalent "mate" somewhere in the list.Consider the set of 6-digit integers, where leading 0's are permitted.
Two integers are considered to be "equivalent" if one can be obtained
from the other by a rearrangement (permutation) of the digits.
Thus 129450 and 051294 are "equivalent".
Among all the 10^6 six-digit integers:
a) how many non-equivalent integers are there?
Example: .000007 has permutations: 000070, 000700, 007000, 070000, 700000
The only six-digit numbers which do not have equivalent mates
. . are those consisting of exactly one digit.
Hence: .000000, 111111, 222222, ..., 999999 are non-equivalent.
Therefore, there are ten non-equivalent integers.
Since 0 and 9 cannot appear six times,b) if digits 0 and 9 can appear at most once, how many non-equivalent integers are there?
. . only: .111111,222222, ..., 888888 are non-equivalent.
Therefore, there are eight non-equivalent integers.
(a) For n-digit numbers, the non-equivalent integers are:c) Generalize your results to (a) and (b) for n-digit integers.
. . . . 000...0, .111...1, .222...2, .. . . , .999...9
. . The answer is still ten.
(b) Similarly, the answer is still eight.