Question:medium

How many different nine-digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

Show Hint

When rearranging digits with repetitions, use the formula for permutations of multiset: \( \frac{n!}{k_1!k_2!...k_m!} \), where \( k_1, k_2, \dots \) are the frequencies of the distinct elements.
Updated On: Jan 13, 2026
  • 16
  • 36
  • 60
  • 100
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: {Digit Classification}
The number comprises the digits \( 2, 2, 3, 3, 5, 5, 8, 8, 8 \). This set includes 4 odd digits (\( 3, 3, 5, 5 \)) and 5 even digits (\( 2, 2, 8, 8, 8 \)). 
Step 2: {Arranging Odd Digits} 
The odd digits must be placed in the even-numbered positions of the nine-digit number. With 4 even positions available, the number of permutations for placing the odd digits is calculated as: \[ \frac{4!}{2!2!} = 6. \] 
Step 3: {Arranging Even Digits} 
The 5 remaining odd positions are to be filled by the even digits. The number of ways to arrange the even digits in these positions is: \[ \frac{5!}{2!3!} = 60. \] Consequently, the total number of possible arrangements is \( 6 \times 60 = 60 \), corresponding to option (C). 
 

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