My understanding of the problem is as follows:
Person A has 100 sets of 10 items because he went to the store 100 times in the year. Now the probability that B goes to the store and buys a set that was also bought by A is 
$$
100*\frac{1}{1000\choose 10}
$$

This was for B's one visit. The 100 that we see in the product comes from the fact that A had 100 items. Now we know B also went to the store 100 times so probability that A and B had a same set of 10 items becomes 
$$
100*\frac{1}{1000\choose 10}*100
$$

Knowing that we have 100 million people, possible pairs of A and B 
$$
{10^8\choose 2}
$$
Hence the final answer to me seems to be 
$$
100*\frac{1}{1000\choose 10}*100*{10^8\choose 2} \sim \frac{10^{20}}{10^{30}}*10\,!
$$
I would be happy to discuss the solution.

The Probability of Catching Criminals
The spreadsheet is Excel 97-2003 compatible and permissions 777

use strict;
use Spreadsheet::ParseExcel;
print "Content-type: text/html\n\n";   
my $parser   = Spreadsheet::ParseExcel-&gt;new();
print "&lt;br&gt;gets here:".__LINE__; 
my $workbook = $parser-&gt;parse('test.xls');
print "&lt;br&gt;never gets here:".__LINE__; 

