Let A = 2, 3, 4, 5, 6.

Question

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

Answer: 120

Accepted Answer

We are given the set $A={2,3,4,5,6}$.

The relation $R$ is defined on the Cartesian product $A	imes A$, meaning its elements are ordered pairs of the form $((x,y),(z,w))$ where all $x,y,z,w\in A$.

The condition for the relation $(x,y)R(z,w)$ is:

$x$ divides $z$

$y\le w$

Notice that the condition on $x$ and $z$ is completely independent of the condition on $y$ and $w$. Therefore, by the Fundamental Principle of Counting, the total number of elements in $R$ will be the number of possible pairs for $(x,z)$ multiplied by the number of possible pairs for $(y,w)$.

Find the number of pairs $(x,z)$ where $x$ divides $z$

Let us list down the possible values for $z\in A$ for each $x\in A$.

If $x=2$, then $z$ can be $2,4,$ or $6$. (This gives 3 pairs)

If $x=3$, then $z$ can be $3$ or $6$. (This gives 2 pairs)

If $x=4$, then $z$ can only be $4$. (This gives 1 pair)

If $x=5$, then $z$ can only be $5$. (This gives 1 pair)

If $x=6$, then $z$ can only be $6$. (This gives 1 pair)

Adding these up, the total number of pairs for $(x,z)$ is:

$3+2+1+1+1=8$

Find the number of pairs $(y,w)$ where $y\le w$

Let us list down the possible values for $w\in A$ for each $y\in A$.

If $y=2$, then $w$ can be $2,3,4,5,$ or $6$. (This gives 5 pairs)

If $y=3$, then $w$ can be $3,4,5,$ or $6$. (This gives 4 pairs)

If $y=4$, then $w$ can be $4,5,$ or $6$. (This gives 3 pairs)

If $y=5$, then $w$ can be $5$ or $6$. (This gives 2 pairs)

If $y=6$, then $w$ can only be $6$. (This gives 1 pair)

Adding these up, the total number of pairs for $(y,w)$ is:

$5+4+3+2+1=15$

Calculate the total number of elements in $R$

Now, we multiply the number of possibilities for $(x,z)$ and $(y,w)$ together:

$R=8	imes 15=120$

Therefore, the number of elements in $R$ is $120$.

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