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(B) The relation $R$ is defined as $R = \{(a, b) : a \in \{1, 2, \ldots, 60\}, b = pq, p, q \geq 3, p, q 	ext{ are primes}, b \leq 60\}$.Since $a$ ca

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$660$

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(B) The relation $R$ is defined as $R = \{(a, b) : a \in \{1, 2, \ldots, 60\}, b = pq, p, q \geq 3, p, q 	ext{ are primes}, b \leq 60\}$.Since $a$ can be any value from $1$ to $60$,there are $60$ choices for $a$.We need to find the number of possible values for $b = pq$ such that $b \leq 60$ and $p, q \geq 3$ are primes.Case $1$: $p = 3$. Then $b = 3q$. Since $b \leq 60$,$3q \leq 60 \implies q \leq 20$. The primes $q \geq 3$ are $3, 5, 7, 11, 13, 17, 19$. There are $7$ values.Case $2$: $p = 5$. Then $b = 5q$. Since $b \leq 60$,$5q \leq 60 \implies q \leq 12$. The primes $q \geq 5$ are $5, 7, 11$. There are $3$ values.Case $3$: $p = 7$. Then $b = 7q$. Since $b \leq 60$,$7q \leq 60 \implies q \leq 8.57$. The prime $q \geq 7$ is $7$. There is $1$ value.Case $4$: $p = 11$. Then $b = 11q$. Since $b \leq 60$,$11q \leq 60 \implies q \leq 5.45$. No prime $q \geq 11$ satisfies this.Total values for $b = 7 + 3 + 1 = 11$.Since $a$ has $60$ choices and $b$ has $11$ choices,the total number of elements in $R$ is $60 	imes 11 = 660$.

Let $R$ be a relation from the set $\{1, 2, 3, \ldots, 60\}$ to itself such that $R = \{(a, b) : b = pq\}$,where $p, q \geq 3$ are prime numbers and $b \leq 60$. Then,the number of elements in $R$ is.

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