Thread: Combination Problem... Help Plz

Question Details:
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?
b. if digits 0 and 9 can appear at most once, how many non-equivalent integers are there?
c. Generalize your results to (a) and (b) for n-digit integers.

My brain is going to explode~~ Thank you for the helps.

Hello, jmmjm!

This seemed like a formidable problem,
. . but with a little thinking . . .

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?

Nearly all six-digit numbers have an equivalent "mate" somewhere in the list.

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.

b) if digits 0 and 9 can appear at most once, how many non-equivalent integers are there?

Since 0 and 9 cannot appear six times,

. . only: .111111,222222, ..., 888888 are non-equivalent.

Therefore, there are eight non-equivalent integers.

c) Generalize your results to (a) and (b) for n-digit integers.

(a) For n-digit numbers, the non-equivalent integers are:
. . . . 000...0, .111...1, .222...2, .. . . , .999...9

. . The answer is still ten.


(b) Similarly, the answer is still eight.