Animated Solution for Mathematics - Permutations and Combinations: The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2, 3, 4, 5, 6 (repetition of digits is not allowed) and divisible by 55 is ______.

Unlocking the Code

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are detectives decoding a sequence. We are tasked with finding natural numbers between $1012$ and $23421$ using the digits $\{2,3,4,5,6\}$ without repetition, such that the number is divisible by $55$.

This is a beautiful puzzle that tests your ability to synthesize number theory with systematic counting.

The Divisibility Filter

Whenever you see a number like $55$, your first instinct should be to break it down into its prime components. Since $55=5\times 11$, and $5$ and $11$ are co-prime, a number is divisible by $55$ if and only if it is independently divisible by both $5$ and $11$.

This is our master key.

For divisibility by $5$, the last digit must be $0$ or $5$. Looking at our available set $\{2,3,4,5,6\}$, we see a glaring absence of $0$.

This is a gift! It means the last digit of every single valid number we construct is fixed as $5$. Our search space has just collapsed from a chaotic mess into a structured hunt.

The 4-Digit Hunt

Let us consider the 4-digit numbers first. They take the form $abc5$. The divisibility rule for $11$ is elegant: the alternating sum of the digits must be a multiple of $11$.

Mathematically, this is:

Here is where the magic happens. We need to bound this expression. The digits $a,b,c$ are chosen from $\{2,3,4,6\}$.

The maximum sum of $a+c$ is $6+4=10$, and the minimum is $2+3=5$. Consequently, the expression $(a+c)-(b+5)$ is trapped in the interval $[-6,3]$.

Within this narrow range, the only multiple of $11$ is $0$. Thus, we must have:

Now, we test our cases for $b$:

If $b=2$, then $a+c=7$. From $\{3,4,6\}$, the only pair summing to $7$ is $\{3,4\}$. This gives us $3245$ and $4235$. If $b=3$, then $a+c=8$. From $\{2,4,6\}$, the only pair summing to $8$ is $\{2,6\}$. This gives us $2365$ and $6325$. If $b=4$, then $a+c=9$. From $\{2,3,6\}$, the only pair summing to $9$ is $\{3,6\}$. This gives us $3465$ and $6435$. If $b=6$, then $a+c=11$. No pair from the remaining digits sums to $11$.

We have found $6$ valid 4-digit numbers. The logic is sound, and the path is clear.

The 5-Digit Mirage

Now, we turn to 5-digit numbers: $abcd5$. The sum of all digits is $2+3+4+6+5=20$.

The alternating sum condition becomes:

Since $b+d=15-(a+c)$, we substitute to get:

Again, $k=0$ is the only solution, leading to $a+c=5$ and $b+d=10$. This forces the pairs $\{a,c\}=\{2,3\}$ and $\{b,d\}=\{4,6\}$.

The smallest number we can form is $24365$. But wait! Our range is $n<23421$.

Since $24365>23421$, these numbers are outside our allowed territory. They are a mirage—mathematically valid for divisibility, but physically invalid for our range.

We have systematically dismantled the problem. We found $6$ valid 4-digit numbers and $0$ valid 5-digit numbers.

The total count is $6$. Remember, in JEE Advanced, it is not just about calculating; it is about checking your constraints at every step. You have done well.