Animated Solution for Mathematics - Permutations and Combinations: A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is

Welcome, future engineers! Today, we are not just solving a permutation problem; we are embarking on a journey of logical deduction. Combinatorics is often called the 'art of counting,' but in the context of the JEE Advanced, it is really the art of constrained counting.

We are given a set of digits: $\{0,1,2,3,4,5\}$, and we need to build a 5-digit number that is divisible by $3$. It sounds simple, but the devil is in the details. Let us break this down, step by step, with the precision of a master architect.

Before we touch a single permutation formula, we must understand the soul of the problem. The ancient rule of arithmetic tells us: a number is divisible by $3$ if and only if the sum of its digits is a multiple of $3$.

Let us look at our toolkit. We have six digits: $\{0,1,2,3,4,5\}$. If we sum them all up, we get $0+1+2+3+4+5=15$.

We are tasked with forming a 5-digit number, which means we must leave exactly one digit behind. Let us call this excluded digit $d$. The sum of our chosen five digits will be $15-d$. For our number to be divisible by $3$, this sum must be a multiple of $3$.

Since $15$ is already a multiple of $3$, the only way for $(15-d)$ to be a multiple of $3$ is if $d$ itself is a multiple of $3$. Looking at our set, the multiples of $3$ are $0$ and $3$. This is our 'Aha!' moment: we have split the problem into two distinct, manageable universes.

The Universe Excluding $0$

If we exclude $0$, our set of digits becomes $\{1,2,3,4,5\}$. The sum is $15-0=15$, which is divisible by $3$.

Here, we have five distinct digits and five slots to fill. Since $0$ is not in this set, we have no 'leading zero' anxiety. Every single permutation of these five digits will result in a valid 5-digit number.

We have secured $120$ valid numbers from this case.

The Trap of the Leading Zero

If we exclude $3$, our set of digits becomes $\{0,1,2,4,5\}$. The sum is $15-3=12$, which is also divisible by $3$.

We have five digits, but one of them is the dreaded $0$. If we blindly calculate $5!$, we will include numbers like $01245$, which is actually a 4-digit number. To solve this, we use the 'Total minus Invalid' strategy.

First, the total permutations ignoring the restriction is $5!=120$. Second, the invalid permutations are those where $0$ sits in the first slot. If we fix $0$ in the first position, we have $4$ remaining slots to fill with the remaining $4$ digits $\{1,2,4,5\}$.

The number of valid arrangements in Case 2 is the total minus the invalid ones:

We have navigated the two cases successfully. We have $120$ ways from the first case and $96$ ways from the second. Since these two cases are mutually exclusive, we simply add them together to find the total number of valid 5-digit numbers.

There you have it! The final answer is $216$.

Remember, in JEE Advanced, the math is rarely the hardest part; it is the logical structure and the ability to spot constraints that separate the top performers. You identified the divisibility rule, partitioned the problem into cases, and navigated the leading-zero trap with grace. Keep this mindset, and no problem will ever be too complex for you to solve.