The given digits are:
\[ \{2, 3, 4, 5, 7\}. \]
A number is divisible by 3 if the sum of its digits is divisible by 3. Identify all cases where the sum of three digits is divisible by 3.
The total number of 3-digit permutations is:
\[ P(5, 3) = 5 \cdot 4 \cdot 3 = 60. \]
Now exclude numbers that are divisible by 3. Compute sums of digits for all groups of three: For digits \( (2, 3, 4), (3, 5, 7) \), etc., find cases where sums like \(2 + 3 + 4 = 9\) (divisible by 3).
Count the total valid cases:
Divisible cases: \(6\) (from permutations of divisible groups).
The remaining numbers are:\[ 60 - 24 = 36. \]
Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option from the following:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): For any two prime numbers $p$ and $q$, their HCF is 1 and LCM is $p + q$.
Reason (R): For any two natural numbers, HCF × LCM = product of numbers.