Question:
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
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
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Arrange restricted digits first, then fill remaining positions and divide by factorials of repeated digits.
Updated On: Jun 3, 2026
- $180$
- $120$
- $60$
- $36$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Arrange digits according to the positional restriction and then account for repeated digits.
Step 2: Meaning
Digits are \[ 2,2,3,3,5,5,8,8,8. \] The odd digits are \[ 3,3,5,5. \] They must occupy the even positions \[ 2,4,6,8. \]
Step 3: Analysis
Number of arrangements of odd digits: \[ \frac{4!}{2!\,2!}=6. \] The remaining positions \[ 1,3,5,7,9 \] are occupied by \[ 2,2,8,8,8. \] Number of arrangements: \[ \frac{5!}{2!\,3!}=10. \] Hence total arrangements: \[ 6\times 10=60. \]
Step 4: Conclusion
Therefore the required number of nine-digit numbers is \[ 60. \]
Final Answer: (C)
Arrange digits according to the positional restriction and then account for repeated digits.
Step 2: Meaning
Digits are \[ 2,2,3,3,5,5,8,8,8. \] The odd digits are \[ 3,3,5,5. \] They must occupy the even positions \[ 2,4,6,8. \]
Step 3: Analysis
Number of arrangements of odd digits: \[ \frac{4!}{2!\,2!}=6. \] The remaining positions \[ 1,3,5,7,9 \] are occupied by \[ 2,2,8,8,8. \] Number of arrangements: \[ \frac{5!}{2!\,3!}=10. \] Hence total arrangements: \[ 6\times 10=60. \]
Step 4: Conclusion
Therefore the required number of nine-digit numbers is \[ 60. \]
Final Answer: (C)
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