The question is from Permutation and Combination. Another question which combines number theory with combinatorics. We need to find out the number of five digits which can be formed with the given conditions. This section hosts a number of questions which are on par with CAT questions in difficulty on CAT Permutation and Combination, and CAT Probability.
Question 8: From the digits 2, 3, 4, 5, 6 and 7, how many 5-digit numbers can be formed that have distinct digits and are multiples of 12?
Any multiple of 12 should be a multiple of 4 and 3.
First, let us look at the constraint for a number being a multiple of 3.
Sum of the digits should be a multiple of 3.
Sum of all numbers from 2 to 7 is 27.
So, if we have to drop a digit and still retain a multiple of 3, we should drop either 3 or 6.
So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.
When the digits are 2, 4, 5, 6, 7.
The last two digits possible for the number to be a multiple of 4 are 24, 64, 52, 72, 56, 76.
For each of these combinations, there are 6 different numbers possible.
So, with this set of 5 digits we can have 36 different numbers.
When the digits are 2, 3, 4, 5, 7. The last two digits possible for the number to be a multiple of 4 are 32, 52, 72, 24.
For each of these combinations, there are 6 different numbers possible.
So, with this set of 5 digits we can have 24 different numbers.
Overall, there are 60 different 5-digit numbers possible.
The question is "how many 5-digit numbers can be formed that have distinct digits and are multiples of 12?"
Choice B is the correct answer.
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