Let S = {p_1, p_2, ldots, p_{10}} be the set | vedclass

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(A) Let $S = \{p_1, p_2, \ldots, p_{10}\}$. The set $P$ consists of all products of distinct elements of $S$. The set $A = S \cup P$ contains all prod

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$5120$

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(A) Let $S = \{p_1, p_2, \ldots, p_{10}\}$. The set $P$ consists of all products of distinct elements of $S$. The set $A = S \cup P$ contains all products of $k$ distinct elements of $S$ for $k = 1, 2, \ldots, 10$. For a fixed $x \in S$,we need to find the number of $y \in A$ such that $x$ divides $y$. If $y = p_{i_1} p_{i_2} \ldots p_{i_k}$,then $x$ divides $y$ if and only if $x \in \{p_{i_1}, \ldots, p_{i_k}\}$. For a fixed $x = p_j$,the number of such products $y$ is the number of subsets of $S$ that contain $p_j$. Since there are $10$ elements in $S$,the number of subsets containing a specific element $p_j$ is $2^{10-1} = 2^9 = 512$. Since there are $10$ choices for $x \in S$,the total number of ordered pairs $(x, y)$ is $10 	imes 512 = 5120$.

Let $S = \{p_1, p_2, \ldots, p_{10}\}$ be the set of the 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 $(x, y)$,where $x \in S$ and $y \in A$,such that $x$ divides $y$,is . . . . . .

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