Let $$A = \{2, 3, 4, 5, 6\}$$. Let $$R$$ be a relation on the set $$A \times A$$ given by $$(x, y)R(z, w)$$ if and only if $$x$$ divides $$z$$ and $$y \le w$$. Then the number of elements in $$R$$ is _________.

We have two ordered pairs $$(x,y),\,(z,w) \in A \times A$$ where $$A=\{2,3,4,5,6\}$$.

For the pair $$\bigl((x,y),(z,w)\bigr)$$ to belong to the relation $$R$$, the two conditions are:
  • $$x$$ divides $$z$$  ($$x \mid z$$)
  • $$y \le w$$.

To count all such pairs it is convenient to treat the two coordinates independently.

1. Counting the choices for $$(x,z)$$ with $$x \mid z$$.
List all elements of $$A$$ and the members of $$A$$ they divide:

  • $$x=2$$ divides $$\{2,4,6\} \;\Rightarrow\; 3\text{ choices for }z$$
  • $$x=3$$ divides $$\{3,6\} \;\Rightarrow\; 2\text{ choices for }z$$
  • $$x=4$$ divides $$\{4\} \;\Rightarrow\; 1\text{ choice for }z$$
  • $$x=5$$ divides $$\{5\} \;\Rightarrow\; 1\text{ choice for }z$$
  • $$x=6$$ divides $$\{6\} \;\Rightarrow\; 1\text{ choice for }z$$

Hence the total number of ordered pairs $$(x,z)$$ satisfying $$x \mid z$$ is $$3+2+1+1+1 = 8.$$ Denote this by $$N_1 = 8.$$

2. Counting the choices for $$(y,w)$$ with $$y \le w$$.
Arrange $$A$$ in ascending order: $$2,3,4,5,6$$. For each possible $$y$$, count the eligible $$w$$:

  • $$y=2$$: $$w$$ can be $$2,3,4,5,6 \;(\!5\text{ choices})$$
  • $$y=3$$: $$w$$ can be $$3,4,5,6 \;(\!4\text{ choices})$$
  • $$y=4$$: $$w$$ can be $$4,5,6 \;(\!3\text{ choices})$$
  • $$y=5$$: $$w$$ can be $$5,6 \;(\!2\text{ choices})$$
  • $$y=6$$: $$w$$ can be $$6 \;(\!1\text{ choice})$$

The total number of ordered pairs $$(y,w)$$ satisfying $$y \le w$$ is $$5+4+3+2+1 = 15.$$ Denote this by $$N_2 = 15.$$

3. Combining the two counts.
The conditions on $$(x,z)$$ and $$(y,w)$$ are independent, so the total number of ordered pairs $$\bigl((x,y),(z,w)\bigr)\in R$$ is the product $$N_1 \times N_2 = 8 \times 15 = 120.$$

Therefore, the relation $$R$$ contains $$\mathbf{120}$$ elements.