Animated Solution for Mathematics - Permutations and Combinations: The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :

Breaking the $6000$ Barrier

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are learning how to build a logical fortress.

When you face a problem like this in the JEE Advanced, the first thing you must do is stop, breathe, and visualize the landscape. We have five distinct digits: $3,5,6,7,$ and $8$. We need to form numbers greater than $6000$ without repeating any digit.

This is a game of constraints, and your job is to identify the rules of the game.

Defining the Search Space

Before we touch a pen to paper, let's look at the length of the numbers we can form. We have five digits available.

Can we form a 1-digit, 2-digit, or 3-digit number greater than $6000$? Absolutely not. The largest 3-digit number we could possibly form is $876$, which is significantly less than $6000$.

So, our search space is immediately narrowed down to two distinct possibilities: 1. 4-digit numbers 2. 5-digit numbers

This is the beauty of combinatorics—by simply analyzing the constraints, we have already simplified the problem into two manageable cases. We will solve each case independently and then bring them together.

The 4-Digit Challenge

Let's tackle the 4-digit numbers first. Imagine four empty slots: $\_\_\_\_$.

The first slot, the thousands place, is the 'gatekeeper'. If we put a $3$ or a $5$ there, the number will be in the $3000$s or $5000$s—both are less than $6000$. To be greater than $6000$, the first digit must be $6,7,$ or $8$.

This gives us exactly $3$ choices for the first slot.

Now, what about the remaining three slots? We started with $5$ digits and used one for the thousands place. That leaves us with $4$ digits.

We need to fill the remaining $3$ slots using these $4$ available digits. This is a classic permutation problem. The second slot has $4$ options, the third has $3$, and the fourth has $2$.

Using the Fundamental Principle of Counting, the total number of 4-digit integers is:

There you have it! We have successfully counted $72$ valid 4-digit numbers.

The 5-Digit Freedom

Now, let's look at the 5-digit numbers. We have five slots to fill: $\_\_\_\_\_$. We have five digits available: $3,5,6,7,8$.

Ask yourself: is there any restriction here? The smallest 5-digit number we can form is $35678$.

Since $35678$ is significantly greater than $6000$, every single 5-digit number we can possibly form will satisfy our condition. There are no 'gatekeepers' here. We are free to arrange all $5$ digits in any order we like.

The number of ways to arrange $5$ distinct items in $5$ slots is simply $5!$ (5 factorial):

The Synthesis

We have reached the final step. We have $72$ valid 4-digit numbers and $120$ valid 5-digit numbers.

Since a number can be a 4-digit number OR a 5-digit number, we use the Addition Principle to combine our results:

And there it is. By breaking the problem into logical, bite-sized pieces, we have navigated the constraints and arrived at the answer: $192$.

Remember, in JEE, the math is rarely the hardest part—it is the clarity of your logic. Keep practicing this systematic approach, and you will find that even the most complex problems start to unravel before your eyes.