How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions ?

Correct Answer: (C) 36

The 8 positions are 1, 2, 3, 4, 5, 6, 7, 8. Odd positions (1, 3, 5, 7) must be occupied by the four odd digits $\{1, 1, 3, 3\}$. The number of distinct permutations for these positions is $\frac{4!}{2!2!}=\frac{24}{4}=6$$\frac{4!}{2! 2!}$ = $\frac{24}{4}$ = 6. Even positions (2, 4, 6, 8) must be occupied by the four even digits $\{2, 2, 4, 4\}$, which likewise have $\frac{4!}{2!2!}=6$$\frac{4!}{2! 2!}$ = 6 distinct permutations. Combining them gives $6$$\times$ 6 = 36 distinct 8-digit numbers.