Hello, jmmjm!

This seemed like a formidable problem,

. . but with a littlethinking. . .

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 dohave equivalent matesnot

. . are those consisting ofexactly one digit.

Hence: .000000, 111111, 222222, ..., 999999 are non-equivalent.

Therefore, there aretennon-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 areeightnon-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 stillten.

(b) Similarly, the answer is stilleight.