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
 A: You are not correct.  The subsets $A_i$ need not be small.  They say come up with the number $n$ such that you cannot avoid having two with the same product.  There are many non-empty subsets of $\{1, 2, \ldots, 9 \}$.  So to prove that 20 is not large enough I could just construct a set of 20 in which no two sets have the same product.  Here is such a sequence in order of increasing product:
$\{ 1 \}$, $\{ 2 \}$, $\{ 3 \}$, $\{ 4 \}$, $\{ 5 \}$, $\{ 6 \}$, $\{ 7 \}$, $\{ 8 \}$, $\{ 9 \}$,
$\{ 2, 5 \}$, $\{ 2, 6\}$, $\{ 2, 7 \}$, $\{ 2, 8 \}$, $\{ 2, 9 \}$, 
$\{ 3, 7 \}$, $\{ 3, 8\}$, $\{ 3, 9 \}$,
$\{ 4, 7 \}$, $\{ 4, 8\}$, $\{ 4, 9 \}$
A different way of looking at the question is: what is the number one larger than the total number of distinct products possible when multiplying distinct numbers between 1 and 9.  
EDIT:
I will provide a solution but in a spoiler block. This is to avoid helping too much in what might be homework:

 We are only counting the number of possible products of the numbers 1 through 9 without repetition.  We can count the distinct products by looking at the prime factorizations of the possible products.  All products $p(A_i)$ will have a prime factorization of the form 

 $$ 2^a 3^b 5^c 7^d $$ 

 where $a$ is in the set $\{0, \ldots, 7\}$, $b$ is in the set $\{0, \ldots, 4\}$, $c$ is in the set $\{0, 1\}$ and $d$ is in the set $\{ 0, 1 \}$.  That makes 160 possibilities.  However the answer is not all possible combinations because the number 6 is either in the product or not, which means that $a$ cannot be 7 when $b$ is 0 and $b$ cannot be 4 when $a$ is 0.  There are eight such forbidden products ($5^{\{0,1\}} 7^{\{0,1\}} 2^0 3^4 $ and $5^{\{0,1\}} 7^{\{0,1\}} 2^7 3^0 $).  All other products are possible.  That leads to 2*2*5*8 - 4*(2) = 152 possible distinct products, $p(A_i)$.  Thus when $n$ is 153 there must be a repetition.  These numbers are small enough that python can quickly brute force the number of distinct products as 152 for extra confirmation.

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