Permutations of 11223344 with positional parity

An 8-digit number contains 4 odd positions (1st, 3rd, 5th, 7th) and 4 even positions (2nd, 4th, 6th, 8th).

Step 1: Arrange the odd digits \{1, 1, 3, 3\} into the 4 odd slots. This is a permutation of items with identical repetitions: Ways = 4!/2! × 2! = 24/4 = 6.

Step 2: Arrange the even digits \{2, 2, 4, 4\} into the 4 even slots. Similarly, this involves identical repetitions: Ways = 4!/2! × 2! = 24/4 = 6.

Since the choices are completely independent, multiply the two independent possibilities together: Total distinct combinations = 6 × 6 = 36.

Answer: (c).