Let $$ S=\{p_1,p_2,....,p_{10}\} $$ be the set of first ten prime numbers. Let $$ A=S\cup P, $$ where $$ P $$ is the set of all possible products of distinct elements of $$ S. $$ Then the number of all ordered pairs $$ (x,y),\; x\in S,\; y\in A, $$ such that $$ x $$ divides $$ y,$$ is: $$ \underline{ \hspace{2cm} } $$

S = set of first 10 primes. A = S ∪ P where P = all possible products of distinct elements of S. Find ordered pairs (x,y) where x∈S, y∈A, x|y.

A consists of S (the 10 primes) plus all products of 2 or more distinct primes from S. Total elements in A = $$2^{10} - 1 = 1023$$ (all non-empty subsets of S, each giving a unique product).

For each prime $$p_i \in S$$, count elements y ∈ A such that $$p_i | y$$.

y is a product of a non-empty subset of S. For $$p_i | y$$, the subset must contain $$p_i$$.

Number of such subsets = $$2^9 = 512$$ (choose any subset of the remaining 9 primes, including empty set, and include $$p_i$$).

Total ordered pairs

Each of the 10 primes in S contributes 512 pairs.

Total = $$10 \times 512 = 5120$$

The answer is 5120.