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 :
Correct Option: 1
Solution:
$\because$ There are total 9 digits and out of which only 3 digits are odd.
$\therefore$ Number of ways to arrange odd digits first
$={ }^{4} C_{3} \cdot \frac{3 !}{2 !}$
Hence, total number of 9 digit numbers
$=\left({ }^{4} C_{3} \cdot \frac{3 !}{2 !}\right) \cdot \frac{6 !}{2 ! 4 !}=180$
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