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?

The number is 11223344. It has 8 digits: four odd digits (1, 1, 3, 3) and four even digits (2, 2, 4, 4). There are 8 positions in the 8-digit number. Odd positions are 1st, 3rd, 5th, 7th (4 positions). Even positions are 2nd, 4th, 6th, 8th (4 positions).

Step 1: Arrange the odd digits (1, 1, 3, 3) in the 4 odd positions. This is a permutation of 4 items with repetitions. The formula is n! / (r1! * r2! * ...), where n is the total number of items, and r1, r2 are the counts of repeated items. Number of ways = 4! / (2! * 2!) = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = 24 / 4 = 6 ways.

Step 2: Arrange the even digits (2, 2, 4, 4) in the 4 even positions. Similarly, this is a permutation of 4 items with repetitions. Number of ways = 4! / (2! * 2!) = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = 24 / 4 = 6 ways.

Step 3: To find the total number of distinct 8-digit numbers, multiply the number of ways for arranging odd digits and even digits, as these are independent choices. Total distinct numbers = (Ways to arrange odd digits) * (Ways to arrange even digits) = 6 * 6 = 36.

Therefore, 36 distinct 8-digit numbers can be formed.

Option Analysis: A) 12: Incorrect. This would be 6+6 or 6*2, not the correct multiplication of independent arrangements. B) 18: Incorrect. C) 36: Correct, as calculated above. D) 72: Incorrect.