Question

How many ten-digit numbers can be formed using all the digits of 2435753228 such that odd digits appear only in even places?

• A. 2! 3! 5!
• B. $(5!)^2$
• C. $\frac{(5!)^2}{3!}$
• D. $\frac{(5!)^2}{3!(2!)^2}$

Answer 1

The Correct Option is D

To solve the problem of how many ten-digit numbers can be formed using all the digits of \(2435753228\) with the restriction that odd digits appear only in even places, we follow these steps: 

1. Identify the individual digits and their frequency from the number \(2435753228\):
   • Odd digits: \(3, 5, 7\)
   • Even digits: \(2, 4, 8\)
2. Count the frequency of each digit:
   • Odd digits: \(3(2), 5(2), 7(1)\) – Total: 5 odd digits
   • Even digits: \(2(3), 4(1), 8(1)\) – Total: 5 even digits
3. Determine the positions for the odd and even digits:
   • Odd digits can occupy even places in a ten-digit number, resulting in positions: 2, 4, 6, 8, 10.
   • Even digits can occupy the remaining positions: 1, 3, 5, 7, 9.
4. Calculate ways to arrange odd digits:
   • We must arrange 5 odd digits \((3, 5, 5, 7)\) in 5 positions. There are duplicates (two '3's and two '5's).
   • Permutation formula considering repetition: \(\frac{5!}{2! \cdot 2!}\).
5. Calculate ways to arrange even digits:
   • Arrange 5 even digits \((2, 2, 2, 4, 8)\) in 5 positions. There are three '2's.
   • Permutation formula considering repetition: \(\frac{5!}{3!}\).
6. Total number of ten-digit numbers is the product of the permutations for odd and even digits:
   • Mathematically, this is: \(N = \frac{5!}{2! \cdot 2!} \times \frac{5!}{3!}\)
7. Simplify the calculation:
   • \(\left( \frac{5! \cdot 5!}{3! \cdot 2! \cdot 2!} \right)\)= \(120 \times 10\)= 1200

The correct answer using the mathematical approach is: \(\frac{(5!)^2}{3!(2!)^2}\)

This corresponds to the option: $\frac{(5!)^2}{3!(2!)^2}$.