Five of Seven

Any product of five numbers of the set can be written by:

$\frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}{a \times b}$

...where $a$ and $b$ are the members of the sets that aren't included in the product.

For any two products to be the same $a \times b$ must be equal for each of them. There are only two such sets of numbers in the set, i.e. $1 \times 6 = 2 \times 3$ and $3 \times 4 = 2 \times 6$ . Hence there re four different products that could've been chosen.

If Alex cannot determine the original numbers, it means that there must be other sets of 5 numbers that result in the same product. This problem can be simplified greatly by realizing that the number of 5-tuples from the set $(1, 2, 3, 4, 5, 6, 7)$ whose product is the same is equal to the number of 2-tuples that satisfy the same conditions, as multiplication is commutative.

We now look for pairs of numbers from this set whose product is the same. This is done quite easily, as there are only $\boxed{4}$ such possibilities. Namely, $(1, 6), (2, 3), (2, 6)$ and $(3, 4)$ . These numbers result is the sets: $(2, 3, 4, 5, 7), (1, 4, 5, 6, 7), (1, 3, 4, 5, 7)$ and $(1, 2, 5, 6, 7)$ respectively.

Since Daniel picked 5 numbers out of 7, it means 2 of them are not chosen. We also know that Alex cannot determine the original product which means that the product of the 5 chosen numbers is the same in all cases. In other words, the product of the 2 remaining numbers is likewise the same in all cases. Since 2x3 = 6x1 and 3x4 = 6x2, there are 4 possible pairs of remaining digits and hence 4 possible sets of 5.

picking five numbers is another way of saying leaving out two so two sets of five integers will be equal if you leave out two pairs with equal product. The only ones that exist are: $1\times 6= 2\times 3$ and $3\times 4=2\times 6$

That's four pairs so the number of distinct sets is $\boxed{4}$