Question:hard

The number of different nine-digit numbers that can be formed by rearranging all the digits of the number $223355888$ so that odd digits always occupy even positions is

Show Hint

Arrange restricted digits first, then fill remaining positions and divide by factorials of repeated digits.
Updated On: Jun 3, 2026
  • $180$
  • $120$
  • $60$
  • $36$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: List the digits and the rule.
The number $223355888$ has digits $2,2,3,3,5,5,8,8,8$. The rule is that odd digits must sit only in even positions.
Step 2: Find odd digits and even slots.
The odd digits are $3,3,5,5$ (four of them). The even positions in a $9$-digit number are positions $2,4,6,8$ (four slots). So the four odd digits fill these four even slots.
Step 3: Arrange the odd digits.
We place $3,3,5,5$ in the four even slots. Because $3$ repeats twice and $5$ repeats twice, the count is \[ \frac{4!}{2!\,2!}=6. \]
Step 4: What is left.
The remaining digits are the even digits $2,2,8,8,8$, and the remaining positions are the odd positions $1,3,5,7,9$ (five slots).
Step 5: Arrange the even digits.
Place $2,2,8,8,8$ in five slots. Here $2$ repeats twice and $8$ repeats three times, so \[ \frac{5!}{2!\,3!}=10. \]
Step 6: Multiply the choices.
Total arrangements $=6\times10=60.$ \[ \boxed{60} \]
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