There are 87 square numbers that use every digit (base 10, no leading zeroes)) exactly once.
(In reply to re(3): Pencil and Paper solution (spoiler)
Yes it does.
But you can get much closer if you extend this parameter and add another one:
a. As to the quantity of qualifying 10- digit numbers 9*9! is totally unacceptable number- the upper bond is definitely below 270451- the number quoted in A036755 (#of 10 last distinct digits of a square -
without "no leading zeroes disclaimer") .
b. Once you divide Charlie's list into 9 sectors, considering the most significant digit,you will find how far from uniform the distribution is:
the shortest interval (starting with digit 3 claims 18 out 87 answers,- while the longest (starting with 1)- only 6.
In each of those intervals there are 9! pandigital numbers , but with different chances to be a square (comment later, below).
As to 67389 "roots" - dense in the beginning - scarcer towards the upper band.
Imagine trying to extrapolate the unknown number ( 87)- having on hand one of the subset's data - no matter what algorithm you apply - you'll get different numbers.
My conclusion - Given 2 sets of integers A of n(A) members and B of n(B)- you can define the bonds for n( intersection set A x B) : generally between 0 and <n(the smaller set).
The result might be close to the ratio n(A)/N(B) for an uniform distribution in both sets -but for a non-uniform only the upper and lower bonds are known .
Our case is not of intersection but of matching - not solvable in general case.
Edited on April 18, 2017, 4:16 am