Question

Total number of six-digit numbers in which all and only odd digits appear is:
${\text{A}}{\text{. }}\dfrac{5}{2}\left( {6!} \right) \\\ {\text{B}}{\text{. 6!}} \\\ {\text{C}}{\text{. }}\dfrac{1}{2}\left( {6!} \right) \\\ {\text{D}}{\text{. None of these}} \\\$

Answer 1

Hint: To find the total number of 6 digit numbers, we list out the number of single digit odd digit numbers. Then we count the number of possibilities.

Complete step-by-step solution -
Single digit odd digits:
1, 3, 5, 7, 9

There are a total of 5 odd digits.

The numbers of spaces in a 6 digit number are 6, _ _ _ _ _ _

Each space can be filled with any of these 5 single digit odd numbers and one among the numbers is repeated.

Clearly, one of the odd digits 1, 3, 5, 7, 9 will be repeated.

The number of selections of the sixth digit is $^5{{\text{C}}_1}$​= 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 is 6! × $\dfrac{5}{2}$​

Then the required number of numbers is $\dfrac{5}{2}$​(6!).
Option A is the correct answer.

Note: The key in solving such types of problems is to identify that one among the numbers is going to be repeated.
The number of ways in which n objects can be arranged is n!
The number of ways in which n objects can be arranged in r different ways is given by $^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r}}!\left( {{\text{n}} - {\text{r}}} \right)!}}$.