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).