Number of different values that obtain from a given set Suppose $S=\{1,2,\dots,7\}$, already we know that, $S$ have $2^7$ subsets. But the problem is, when we multiply elements of each subset $S$, how many different value we can obtain from multiplication of elements of each subsets? In $2^7$ subsets, some of them are equal, for example, for subset of size 2, $3\times 4=6\times 2$. Are there a simple way that eliminate the duplicate values for any subsets?
 A: $7!=2^4\cdot 3^2\cdot 5^1\cdot 7^1$ has $(4+1)(2+1)(1+1)(1+1)=60$ divisors
Of these, the divisors who are multiples of $2^4$ but not multiples of $3$ are not possible to make (as $6$ would have been needed for that last factor of $2$ and would have brought a factor of $3$ along with it) as well as the divisors who are multiples of $9$ but not multiples of $2$ are not possible (for a similar reason).
There are $2\cdot 2 = 4$ factors who are multiples of $2^4$ but not multiples of $3$ (seen by looking at factors of the form $2^4\cdot 3^0\cdot 5^z\cdot 7^w$) and similarly there are $2\cdot 2 = 4$ factors who are multiples of $9$ but not multiples of $2$ (looking at $2^0\cdot 3^2\cdot 5^z\cdot 7^w$).
As such the total will be
$$60 - 4-4 = 52$$
A: Since $1$ has no effect in the multiplications you can put it aside. Put the number $6$ aside for now. Factorize all the other numbers. The multiplication of the elements of any nonempty subset of $\lbrace 2, 3, 4, 5, 7 \rbrace$ would be of the form
$$2^{x} \times 3^{y} \times 5^{z} \times 7^{w} $$ where $x$ and $y$ and $z$ and $w$ are whole numbers and $0 \leq x \leq 3$ and $0 \leq y \leq 1$ and $0 \leq z \leq 1$ and $0 \leq w \leq 1$. As a result of the multiplication principle, you can say that there are $4 \times 2 \times 2 \times 2 = 32$ such multiplications. Now what if we play the number $6$ into the game? Then any multiplication would be of the form
$$2^{x + 1} \times 3^{y + 1} \times 5^{z} \times 7^{w} $$ where $x$ and $y$ and $z$ and $w$ are whole numbers and $0 \leq x \leq 3$ and $0 \leq y \leq 1$ and $0 \leq z \leq 1$ and $0 \leq w \leq 1$. However, if $0 \leq x \leq 2$ and $y = 0$ no new state will be generated and vice versa so you should suppose either $x = 3$ or $y = 1$. You can deduce from the inclusion-exclusion principle that the number of the new states will be $2 \times 2 \times 2 + 4 \times 2 \times 2 \; - 2 \times 2 = 20$. So overall the answer would be $32 + 20= 52$.
