Answer :
6, is 24 × 2 = 48
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To form a five-digit number divisible by 6, it must fulfill two conditions: divisibility by 2 and divisibility by 3.
Divisibility by 2: The number must end in an even digit. From the given digits {1, 3, 2, 5, 7}, we have two even digits, which are 2 and 5.
Divisibility by 3: The sum of the digits must be divisible by 3. The sum of the given digits is 1 + 3 + 2 + 5 + 7 = 18, which is divisible by 3. This condition is met for any permutation of the given digits.
Now, we have to find the number of permutations of the remaining digits (excluding the even digit already chosen for the units place) to form the remaining four digits of the five-digit number.
For the thousands place, there are 4 remaining digits.
For the hundreds place, there are 3 remaining digits.
For the tens place, there are 2 remaining digits.
For the units place, we have already chosen one even digit.
So, the total number of permutations for the remaining four digits is 4 × 3 × 2 = 24.
However, we have to account for the fact that the number can end in either 2 or 5. Hence, we multiply the permutations by 2.
Therefore, the total number of five-digit numbers that can be formed, which are divisible by 6, is 24 × 2 = 48.
