How many different nine-digit numbers can be formed from the digits of $223355888$ by rearranging its digits so that the odd digits occupy even positions?
Step-by-Step Solution
Key Concept: Odd digits: $3, 3, 5, 5$ (4 digits) must fill even positions $2, 4, 6, 8$ (4 slots): $\frac{4!}{2! 2!} = 6$ ways. Even digits: $2, 2, 8, 8, 8$ fill odd positions $1, 3, 5, 7, 9$ (5 slots): $\frac{5!}{2! 3!} = 10$ ways. Total $= 6 \times 10 = 60$.
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Correct Answer: (1)