Animated Solution for Mathematics - Permutations and Combinations: 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 Combinatorial Journey

Welcome, my dear student. Today, we are going to unravel a beautiful puzzle in combinatorics. We have been given the digits of the number $223355888$ and asked to rearrange them into a new $9$-digit number, but with a twist: the odd digits must occupy the even positions.

This is not just a math problem; it is a choreography of numbers. Let us break it down step by step.

Deconstruction

First, we must understand our cast of characters. We have the digits $2,2,3,3,5,5,8,8,8$.

Let us categorize them by their parity. The odd digits are $3,3,5,5$, giving us a total of $4$ odd digits. The even digits are $2,2,8,8,8$, giving us a total of $5$ even digits.

This separation is crucial because the problem imposes a specific rule on where these groups can live.

The Constraint

Imagine $9$ empty chairs lined up, numbered $1$ through $9$. The problem tells us that the odd digits must sit in the even-numbered chairs: $2,4,6,$ and $8$.

If you look closely, there are exactly $4$ such chairs. And look at that—we have exactly $4$ odd digits! It is a perfect match.

Now, what about the even digits? They must occupy the remaining chairs: $1,3,5,7,$ and $9$. There are $5$ such chairs, and we have $5$ even digits. The stage is set.

The Odd Arrangement

Let us focus on the odd digits first. We need to arrange $3,3,5,5$ into $4$ positions. If all these digits were unique, we would have $4!$ ways.

But they are not unique! The digit $3$ repeats twice, and the digit $5$ repeats twice. When we have identical items, we must divide by the factorial of their frequencies to avoid overcounting.

The formula is $\frac{n!}{p!\times q!}$, where $n=4$, $p=2$ (for the two $3$s), and $q=2$ (for the two $5$s). So, we calculate:

There are $6$ distinct ways to arrange our odd digits.

The Even Arrangement

Now, let us turn our attention to the even digits: $2,2,8,8,8$. We have $5$ positions to fill. Again, we use the permutation formula for multisets.

We have $5$ digits in total, so the numerator is $5!$. The digit $2$ repeats twice, and the digit $8$ repeats three times. Our calculation becomes:

There are $10$ distinct ways to arrange the even digits.

The Synthesis

We are almost there. We have $6$ ways to arrange the odd digits and $10$ ways to arrange the even digits.

Because these two arrangements are independent—choosing one arrangement for the odd digits does not restrict our choices for the even digits—we use the Fundamental Principle of Counting. We multiply the two results:

And there you have it! There are exactly $60$ different $9$-digit numbers that satisfy our condition. I hope you can see the elegance in this. We didn't just calculate; we organized the chaos into a structured, logical flow. Keep practicing, and soon, these patterns will become second nature to you.