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?

Question

Accepted Answer

The correct answer is option C. The 8-digit number has 8 positions. The odd positions are 1st, 3rd, 5th, 7th (4 positions). The even positions are 2nd, 4th, 6th, 8th (4 positions).

The given digits are 1, 1, 2, 2, 3, 3, 4, 4.
Odd digits: 1, 1, 3, 3 (4 digits)
Even digits: 2, 2, 4, 4 (4 digits)

First, arrange the odd digits (1, 1, 3, 3) in the 4 odd positions. The number of ways to arrange these digits with repetitions is given by 4! / (2! × 2!) = (4 × 3 × 2 × 1) / ((2 × 1) × (2 × 1)) = 24 / 4 = 6 ways.

Next, arrange the even digits (2, 2, 4, 4) in the 4 even positions. Similarly, the number of ways is 4! / (2! × 2!) = 24 / 4 = 6 ways.

To find the total number of distinct 8-digit numbers, we multiply the number of ways to arrange the odd digits by the number of ways to arrange the even digits:
Total ways = 6 × 6 = 36.
Therefore, 36 distinct 8-digit numbers can be formed.

Q29. 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?

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