Question

All possible numbers are formed using the digits $1,1,2,2,2,2,3,4,4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is:

• A. $160$
• B. $162$
• C. $175$
• D. $180$

Answer 1

The correct option is D $180$

There are total $9$ digits, out of which only $3$ are odd.
Also, in a $9$ digit number, there are $4$ even places.
So, these $3$ odd digits will occupy $3$ places out of $4$ and rest all the digits will occupy remaining $6$ places.
Hence, required number of such numbers will be:
${}^4{C}_3.\frac{3!}{2!}.\frac{6!}{4!2!}=180$