Question

If repetition of digits is allowed, how many five-digit numbers less than 13000 are possible such that the product of the digits of the number is 96?

• A. 24
• B. 30
• C. 36
• D. 42
• E. 46

Answer 1

The Correct Option is C

\(96 = 2^5 \times 3\)

Possible \(5\)-digit numbers less than \(13000\):

\(11268, 11348, 11446, 12238, 12246, 12344\)

For \(11268\), if the number is of the form \(11xxx\), we can arrange \(2, 6\) and \(8\) in \(3! = 6\) ways

If the number is of the form \(12xxx\), we can arrange \(1, 6\) and \(8\) in \(3! = 6\) ways

So, \(11268\) can be arranged in \(3! + 3! = 6 + 6 = 12\) ways.

For \(11348\), the number has to be of the form \(11xxx\)

So, \(3, 4\) and \(8\) can be arranged in \(3! = 6\) ways

For \(11446\), the number has to be of the form \(11xxx\)

So, \(4, 4\) and \(6\) can be arranged in = \(\frac{3!}{2!} = 3\) ways

For \(12238\), the number has to be of the form \(12xxx\)

So, \(2, 3\) and \(8\) can be arranged in \(3!\) ways = \(6\) ways

Similarly, \(12246\) can be arranged in \(3!\) ways = \(6\) ways

And \(12344\) can be arranged in = \(\frac{3!}{2!} = 3\) ways

Thus, total number of \(5\)-digit numbers = \(12 + 6 + 3 + 6 + 6 + 3 = 36\)

Hence, option C is the correct answer.