Question

The number of $6$-digit numbers that can be made with the digits $0,1,2,3,4,5$, so that even digits occupy odd places, is

• A. $24$
• B. $36$
• C. $48$
• D. None of these

Answer 1

The correct option is A $24$

In $1st$ digit, only even digits $(0,2,4)$ are to be placed and $0$ cannot be placed at first digit else it will become $5$ digit number.

So, number of digits that can be filled $=2$
In $2nd$ digit, only odd digits $(1,3,5)$ are to be placed, so number of digits that can be filled $=3$
Now, at $3rd$ place, $2$ even digits are left so number of digits that can be filled $=2$
Now, at $4th$ place, $2$ odd digits are left so number of digits that can be filled $=2$
Now, at $5th$ place, $1$ even digit is left so number of digits that can be filled $=1$
Now, at $6th$ place, $1$ odd digit is left so number of digits that can be filled $=1$

So, Total numbers that can be formed $=2\times 3\times 2\times 2\times 1\times 1=24$

Therefore, answer is $24$.