Animated Solution for Mathematics - Permutations and Combinations: Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is

Welcome, future engineer! Today, we are going to peel back the layers of a classic combinatorics problem. It might seem like a simple counting exercise, but it is actually a masterclass in understanding constraints and the Fundamental Principle of Counting.

Imagine you are tasked with building a four-digit number using the digits $\{0,1,2,3,4\}$. The catch is that the number must be greater than $1000$ and less than $4000$. Let us break this down step by step.

The Gatekeeper (The Thousands Place)

In any range-based counting problem, the first step is to identify the most restricted position. Here, that is the thousands place. It acts as the gatekeeper for our range constraint, $1000<N<4000$.

If we place a $0$ in the thousands place, the number effectively becomes a three-digit number, which is less than $1000$. If we place a $4$ in the thousands place, the smallest number we could form is $4000$, which violates our condition that the number must be strictly less than $4000$.

Therefore, the thousands place can only be occupied by $1$, $2$, or $3$. This gives us exactly $3$ valid choices for the first position.

The Freedom of the Remaining Digits

Now, let us look at the hundreds, tens, and units places. The problem states that repetition is allowed. This means that once we have locked in our thousands place, the remaining positions are completely free.

For the hundreds place, we have the full set of available digits: $\{0,1,2,3,4\}$. That is $5$ choices.

The same logic applies to the tens place and the units place. Each of these positions has $5$ available options, as each position is an independent event.

The Fundamental Principle of Counting

Now, we arrive at the heart of the solution: the Fundamental Principle of Counting. This principle states that if one task can be done in $m$ ways and a second task can be done in $n$ ways, then the two tasks together can be done in $m\times n$ ways.

We have four tasks: filling the thousands, hundreds, tens, and units places. We have $3$ choices for the first, $5$ for the second, $5$ for the third, and $5$ for the fourth. To find the total number of valid combinations, we calculate:

Let us perform the final arithmetic. Since $5\times 5\times 5=125$, multiplying this by $3$ gives us $375$.

You have successfully navigated the constraints. Keep this logic in your toolkit; it will serve you well in the most complex JEE problems.