Question

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?

• A. 16
• B. 36
• C. 60
• D. 100

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.

Answer 1

The Correct Option is C

Step 1: {Odd and even digits} 
The digits of the number are \( 2, 2, 3, 3, 5, 5, 8, 8, 8 \). We have 4 odd digits \( 3, 3, 5, 5 \) and 5 even digits \( 2, 2, 8, 8, 8 \). 
Step 2: {Placing odd digits in even positions} 
We need to place the odd digits in the even positions of the nine-digit number. There are 4 even positions, and the number of ways to place the odd digits is: \[ \frac{4!}{2!2!} = 6. \] 
Step 3: {Placing even digits in odd positions} 
There are 5 odd positions remaining for the even digits. The number of ways to place the even digits is: \[ \frac{5!}{2!3!} = 60. \] Thus, the total number of ways is \( 6 \times 60 = 60 \), which matches option (C). 
 

Answer 2

Step 1: Identify digits and their counts
The given number is 223355888.
Digits: 2 (twice), 3 (twice), 5 (twice), 8 (three times).
Odd digits: 3, 3, 5, 5 (4 odd digits total)
Even digits: 2, 2, 8, 8, 8 (5 even digits total)

Step 2: Positions in the 9-digit number
Positions are numbered 1 to 9.
Even positions are 2, 4, 6, 8 — total 4 positions.
Odd positions are 1, 3, 5, 7, 9 — total 5 positions.

Step 3: Condition — odd digits in even positions
All 4 odd digits must be placed in the 4 even positions.
All 5 even digits must be placed in the 5 odd positions.

Step 4: Arrange odd digits (3,3,5,5) in 4 even positions
Number of ways = permutations of 4 digits with duplicates:
\[ \frac{4!}{2! \times 2!} = \frac{24}{4} = 6 \]

Step 5: Arrange even digits (2,2,8,8,8) in 5 odd positions
Number of ways = permutations of 5 digits with duplicates:
\[ \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = \frac{120}{12} = 10 \]

Step 6: Total number of different nine-digit numbers
Multiply the number of ways for odd and even digits:
\(6 \times 10 = 60\)

Final Answer: 60