Animated Solution for Mathematics - Permutations and Combinations: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is

A Journey Through Constraints

Welcome, future engineer. Today, we aren't just solving a counting problem; we are building a logical structure. When you face a problem like this in the JEE Advanced, the temptation is to start scribbling numbers immediately. Resist that urge.

Instead, pause and visualize the constraints. We are tasked with forming 3-digit numbers using the set $\{1,2,3,4,5\}$ without repetition, such that the resulting number is divisible by either 3 or 5. This is a classic test of your ability to manage overlapping conditions.

Our Logical Compass

Whenever you see the word 'OR' in a counting problem, your brain should immediately trigger the Principle of Inclusion-Exclusion. We are looking for the size of the union of two sets: those divisible by 3 and those divisible by 5.

The formula is elegant and powerful:

This formula is our safety net. It ensures that we don't double-count the numbers that are divisible by both 3 and 5—the multiples of 15. Let’s break this down piece by piece.

The Divisibility by 5

Divisibility by 5 is a 'positional' constraint. A number is divisible by 5 if its last digit is 0 or 5. Looking at our set $\{1,2,3,4,5\}$, we see that 0 is absent.

Thus, the units place is locked: it must be 5. With the units place fixed, we have two spots left (hundreds and tens) and four digits remaining $\{1,2,3,4\}$. The number of ways to fill these two spots is a simple permutation problem:

We have found 12 numbers divisible by 5. Simple, clean, and effective.

The Divisibility by 3

Divisibility by 3 is a 'summation' constraint. A number is divisible by 3 if and only if the sum of its digits is a multiple of 3. We need to find all subsets of 3 digits from our set that satisfy this.

Let's be systematic:

1. $\{1,2,3\}$ (Sum = 6) 2. $\{1,3,5\}$ (Sum = 9) 3. $\{2,3,4\}$ (Sum = 9) 4. $\{3,4,5\}$ (Sum = 12)

Each of these 4 subsets can be arranged in $3!=6$ ways. Therefore, the total count for $n(3)$ is:

The Intersection (The Multiples of 15)

Now, we must identify the numbers that were counted in both groups. These are the numbers that end in 5 AND have a digit sum divisible by 3. Looking back at our valid subsets, which ones contain the digit 5? Only $\{1,3,5\}$ and $\{3,4,5\}$.

For the subset $\{1,3,5\}$, if the last digit is 5, the remaining two digits (1 and 3) can be arranged in $2!=2$ ways (135 and 315). Similarly, for $\{3,4,5\}$, the remaining digits (3 and 4) can be arranged in $2!=2$ ways (345 and 435).

Thus:

We have all our components. We have the count of numbers divisible by 3, the count of those divisible by 5, and we have identified the overlap that we must remove to avoid double-counting.

Applying our formula one last time:

There it is. 32 unique numbers. You see, the beauty of this problem isn't in the arithmetic; it's in the organization. By breaking a complex 'OR' condition into manageable, distinct pieces, you turned a chaotic problem into a clear, logical path. Keep this discipline, and no problem will ever be too complex for you to solve.