Let A = 2, 3, 4, 5, 6. Let R be a relation on the set A × A given by (x, y) R (z, w) if and only if x divides z and y ≤ w. Then the number of…

Given the set $A = \{2, 3, 4, 5, 6\}$A={2,3,4,5,6}.

The relation $R$R is defined on $A \times A$A×A such that $(x, y) R (z, w)$(x,y)R(z,w) if and only if $x$x divides $z$z and $y \leq w$y≤w.

The number of elements in $R$R is the number of ordered pairs $((x, y), (z, w))$((x,y),(z,w)) satisfying these conditions. Since the conditions for $(x, z)$(x,z) and $(y, w)$(y,w) are independent, the total number of elements in $R$R will be the product of the number of pairs $(x, z)$(x,z) satisfying $x$x divides $z$z and the number of pairs $(y, w)$(y,w) satisfying $y \leq w$y≤w.

First, we find the number of pairs $(x, z) \in A \times A$(x,z)∈A×A such that $x$x divides $z$z:

For $x = 2$x=2, $z \in \{2, 4, 6\}$z∈{2,4,6} ($3$3 pairs)

For $x = 3$x=3, $z \in \{3, 6\}$z∈{3,6} ($2$2 pairs)

For $x = 4$x=4, $z \in \{4\}$z∈{4} ($1$1 pair)

For $x = 5$x=5, $z \in \{5\}$z∈{5} ($1$1 pair)

For $x = 6$x=6, $z \in \{6\}$z∈{6} ($1$1 pair)

Total number of pairs $(x, z) = 3 + 2 + 1 + 1 + 1 = 8$(x,z)=3+2+1+1+1=8.

Next, we find the number of pairs $(y, w) \in A \times A$(y,w)∈A×A such that $y \leq w$y≤w:

For $y = 2$y=2, $w \in \{2, 3, 4, 5, 6\}$w∈{2,3,4,5,6} ($5$5 pairs)

For $y = 3$y=3, $w \in \{3, 4, 5, 6\}$w∈{3,4,5,6} ($4$4 pairs)

For $y = 4$y=4, $w \in \{4, 5, 6\}$w∈{4,5,6} ($3$3 pairs)

For $y = 5$y=5, $w \in \{5, 6\}$w∈{5,6} ($2$2 pairs)

For $y = 6$y=6, $w \in \{6\}$w∈{6} ($1$1 pair)

Total number of pairs $(y, w) = 5 + 4 + 3 + 2 + 1 = 15$(y,w)=5+4+3+2+1=15.

The total number of elements in the relation $R$R is the product of the number of these pairs:

Total elements = $8 \times 15 = 120$8×15=120.

Answer: $120$120