Question

Total number of six-digit numbers in which all and only odd digits appear is

• A. $\frac{5}{2}(6!)$
• B. $6!$
• C. $\frac{1}{2}(6!)$
• D. None of these

Answer 1

The correct option is A $\frac{5}{2}(6!)$

Clearly, one of the odd digits $1,3,5,7,9$ will be repeated.
The number of selections of the sixth digit is ${}^5{C}_1=5$

As only one digit is repeated so we have $5$ ways in which we can choose that digit, and as it is a repeated number so we are dividing the result by $2$ and since $6$ digits are there so we can arrange them in $6!$ ways.

Now the result for the arrangement for $6$ digits are $6!\times \frac{5}{2}$

Then the required number of numbers is $\frac{5}{2}(6!)$.