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 :
The correct answer is D:180 Given that: The periodic number to be formed using digits \(1,1,2,2,2,2,2,3,4,4\) to be taken all at a time. [Note: Means;it is a clear case of combination as only selection is important here] \(\therefore 4_{C_{3}}\times\frac{3!}{2!}\times\frac{6!}{2!\times4!}=180\)\((\therefore n_{c_{k}}=\frac{n!}{(n-k)!,k!})\)\((4_{c_{3}}=\frac{4!}{1!\times3!})\)