Animated Solution for Mathematics - Permutations and Combinations: Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1, 2, 3, 4, 5 and 6 without repetition of digits. Then the total number of such numbers is ______.

Mastering Constraints

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are learning the art of systematic thinking. In the JEE Advanced, the difference between a correct answer and a trap-ridden mistake often lies in how you organize your thoughts.

This problem asks us to form numbers between $1000$ and $3000$ using the set $\{1,2,3,4,5,6\}$ without repetition, such that the number is divisible by $4$. Let us break this down.

Defining the Boundaries

Imagine the number as a sequence of four slots: $d_1{d}_2{d}_3{d}_4$. The constraint $1000<N<3000$ is our first anchor.

It tells us immediately that the thousands digit, $d_1$, is restricted. It cannot be $3,4,5,$ or $6$, because those would create numbers $\ge 3000$. It cannot be $0$ (not in our set).

Thus, $d_1$ must be either $1$ or $2$. We will solve for $d_1=1$ and $d_1=2$ separately, and then sum them up.

The Divisibility Rule

A number is divisible by $4$ if and only if the number formed by its last two digits, $d_3{d}_4$, is divisible by $4$. This is our second anchor.

We need to find all possible pairs from our set $\{1,2,3,4,5,6\}$ that satisfy this. The 'no repetition' rule means the available digits for $d_3{d}_4$ depend on what we chose for $d_1$.

Case I - The Thousands Digit is $1$

If $d_1=1$, our available digits for the remaining positions are $\{2,3,4,5,6\}$. We look for pairs from this set that are divisible by $4$:

- $24,32,36,52,56,64$

There are $6$ valid pairs. For the hundreds place $d_2$, we have used $d_1$ (which is $1$) and two digits for the pair $d_3{d}_4$.

That is $3$ digits used out of $6$, leaving $6-3=3$ digits remaining. Thus, there are $3$ choices for $d_2$. The total for Case I is:

Case II - The Thousands Digit is $2$

If $d_1=2$, our available digits for the remaining positions are $\{1,3,4,5,6\}$. We must re-evaluate the valid pairs for $d_3{d}_4$ using this new set:

- $16,36,56,64$

Note that $24,32,$ and $52$ are excluded because they require the digit $2$, which is already occupied in the thousands place. We are left with $4$ valid pairs.

Again, we have used $3$ digits total ($d_1$ and the pair $d_3{d}_4$), leaving $3$ choices for $d_2$. The total for Case II is:

We have $18$ numbers from Case I and $12$ numbers from Case II. Since these cases are mutually exclusive, we simply add them:

We have built a logical structure by respecting the constraints and handling the 'no repetition' rule by updating our available digits. The final answer is 30.