Question on Sets Given a finite nonempty set $S$ of integers, let $p(S)$ denote the
product of all integers in $S$ (for example, if $S = \{3, 11, 61\}$,
then $p(S) = 2013$). Determine the least positive integer $n$ such
that in any sequence $A_1, A_2, \cdots A_n$ of $n$ nonempty subsets of
$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, there exist sets $A_i
,A_j$ with $i\not=j$ and $p(A_i) = p(A_j)$.
I have made an attempt to answer the problem.  Could someone let me know if this is correct?
The way I have the arranged the solution is to find the equivalence of factors that might satisfy $p(A_i) = p(A_j)$
If $A_i$ = {1, any combination of from 2 to 9} is equivalent to $A_j$ = { similar combination of from 2 to 9}
= 2*(${8\choose1} +{8\choose2} + \cdots + {8\choose8})$ = $2*(2^{8}-1)$ = 510
Next would be the equivalence of 2*3*Any combination of 4,5,7,8,9 = 6*Any combination of (4,5,7,8,9)
=  2*(${5\choose1} +{5\choose2} + \cdots + {5\choose5})$ = $2*(2^{5}-1)$ = 62
Next would be the equivalence of 2*4*Any combination of 3,5,6,7,9  = 8*Any combination of 3,5,6,7,9 
= 2* (${5\choose1} +{5\choose2} + \cdots + {5\choose5})$ = $2*(2^{5}-1)$ = 62
Next would be the equivalence of  2*9*Any combination of 5,6,7,8 = 3*6*Any combination of (5,7,8,9)
= 2*(${4\choose1} +{4\choose2} + \cdots + {4\choose4})$ = $2*(2^{4}-1)$ = 30
Some double counting among equivalence would 2*3*4*(any combination of 5,7,9)
= (${3\choose0} +{3\choose1} +\cdots + {3\choose3})$  = $(2^{3})$ = 8
The least n would then be
= 510-62-62-30+8 = 364

This is not an answer. The following examples show that $n$ has to be larger than $48$.
$$\{1\},\{2\},\cdots,\{9\}$$
$$\{2,5\},\{2,6\},\{2,7\},\{2,8\},\{2,9\},\{3,7\},\{3,8\},\{3,9\},\{4,7\},\{4,8\},\{4,9\},\{5,8\},\{5,9\},\{6,8\},\{6,9\},\{7,8\},\{7,9\},\{8,9\}$$
$$\{2,8,9\},\{3,8,9\},\cdots,\{7,8,9\}$$
$$\{2,7,8,9\},\{3,7,8,9\},\cdots,\{6,7,8,9\}$$
$$\{2,6,7,8,9\},\{3,6,7,8,9\},\cdots, \{5,6,7,8,9\}$$
$$\{2,5,6,7,8,9\},\{3,5,6,7,8,9\},\{4,5,6,7,8,9\}$$
$$\{2,4,5,6,7,8,9\},\{3,4,5,6,7,8,9\}$$
$$\{2,3,4,5,6,7,8,9\}$$
