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?

EXPLANATION

Correct Option (3)

The given 8-digit number is 11223344. This number contains four odd digits (two 1s and two 3s) and four even digits (two 2s and two 4s).

According to the condition, odd digits must occupy the odd positions (1st, 3rd, 5th, 7th) and even digits must occupy the even positions (2nd, 4th, 6th, 8th).

The number of ways to arrange the four odd digits (1, 1, 3, 3) in the four odd positions is calculated using the formula for permutations with repetitions: $\frac{4!}{2! \times 2!} = \frac{24}{2 \times 2} = \frac{24}{4} = 6$ .
Similarly, the number of ways to arrange the four even digits (2, 2, 4, 4) in the four even positions is also: $\frac{4!}{2! \times 2!} = \frac{24}{2 \times 2} = \frac{24}{4} = 6$ .

Since the arrangement of odd digits in odd positions and even digits in even positions are independent events, the total number of distinct 8-digit numbers is the product of these two possibilities: $6 \times 6 = 36$ .

Incorrect Options:

Options 1 (12), 2 (18), and 4 (72) are incorrect as they do not correspond to the result obtained by correctly applying the principles of permutations for the given constraints.