Animated Solution for Mathematics - Permutations and Combinations: The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is

A Combinatorial Journey

Welcome, future engineers! Today, we are going to dive into the elegant world of combinatorics. Often, students look at a problem like this and feel overwhelmed by the sheer number of possibilities.

But remember, mathematics is not about brute force; it is about finding the hidden structure in the chaos. Let us break down this problem step by step.

Defining the Boundaries

We are tasked with finding how many numbers exist strictly between $5000$ and $10000$ using the digits $\{1,3,5,7,9\}$ without repetition.

The first step in any counting problem is to define the 'universe' of our numbers. Since the range is strictly between $5000$ and $10000$, we immediately realize that we are only interested in $4$-digit numbers.

A $3$-digit number would be too small, and $10000$ is a $5$-digit number, which is outside our upper bound. So, our goal is to fill four slots: Thousands, Hundreds, Tens, and Units.

The Gatekeeper

Now, let us look at the Thousands place. This is our 'gatekeeper.'

Why? Because it dictates whether our number satisfies the condition of being greater than $5000$. If we place a $1$ or a $3$ here, the number will be in the $1000$s or $3000$s, which fails our condition.

Therefore, the only digits from our set $\{1,3,5,7,9\}$ that can occupy the thousands place are $\{5,7,9\}$. This gives us exactly $3$ valid choices for the first slot.

The Cascade of Choices

With the gatekeeper position filled, we move to the remaining slots. We have used one digit out of our set of five.

The constraint is 'no repetition,' which means that digit is now removed from our pool. We are left with $4$ digits for the hundreds place.

Since there are no further restrictions on the hundreds, tens, or units places, we simply continue the countdown. For the hundreds place, we have $4$ choices. For the tens place, having used two digits already, we have $3$ choices remaining. Finally, for the units place, we are left with $2$ choices.

The Power of Multiplication

This brings us to the Fundamental Principle of Counting. We have $3$ choices for the thousands, $4$ for the hundreds, $3$ for the tens, and $2$ for the units.

To find the total number of valid combinations, we multiply these independent choices:

Calculating this, we get:

It is truly beautiful how a complex-sounding constraint simplifies into a clear, logical sequence of choices. Keep practicing this systematic approach, and you will find that even the most daunting combinatorics problems become a joy to solve.

The total number of valid integers is $72$. Happy studying!