The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3,

Q: Wait, why do we only check $O - E = 11$ and $O - E = -11$? Could it be $22$ or $0$?

Excellent question! We must respect the constraints of our digits. The maximum sum of any $3$ digits is $9+7+5 = 21$, and the minimum is $1+2+3 = 6$. Since $E$ must be between $6$ and $21$, and $O$ is the sum of the remaining $4$ digits, the difference $O - E$ is strictly bounded. Calculating these bounds reveals that only $\pm 11$ are mathematically possible.

Q: Why do we multiply by $3!$ and $4!$ at the end?

Think of the slots as distinct entities. Once you choose the set of digits for the even positions, you can shuffle them in $3!$ ways. Simultaneously, the remaining digits for the odd positions can be shuffled in $4!$ ways. Since these are independent choices for a single valid configuration, we multiply them using the Fundamental Principle of Counting.

Q: Is there a trick to finding the sets that sum to $10$ or $21$?

The trick is systematic exhaustion! Start with the largest numbers to reach $21$ quickly, and for $10$, start with the smallest numbers. Don't just guess; organize your search by fixing one digit and finding the pairs that complete the sum. This ensures you never miss a combination.

Question

The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, warriors of JEE. Today, we are not just solving a permutation problem; we are embarking on a journey into the heart of number theory.

We have seven digits: $\{1, 2, 3, 4, 5, 7, 9\}$. We need to arrange them into a $7$-digit number that is a multiple of $11$.

The sum of our digits is $S = 1+2+3+4+5+7+9 = 31$. This sum serves as our mathematical anchor.

### The Alternating Balance

Recall the divisibility rule for $11$: a number is divisible by $11$ if the alternating sum of its digits is a multiple of $11$.

Let $O$ be the sum of the digits in the four odd positions, and $E$ be the sum of the digits in the three even positions. The rule states that:

$$|O - E| = 11k$$

This is the core reality of our problem: we are splitting our set of digits into two teams, the 'Odd Team' and the 'Even Team', such that their difference is a multiple of $11$.

### The Algebraic Constraint

We have two governing equations: $O + E = 31$ and $O - E = 11k$.

Since $O + E = 31$ is odd, $O - E$ must also be odd. This forces $11k$ to be odd, meaning $k$ must be an odd integer.

Given the range of our digits, the only possible values for $O - E$ are $11$ and $-11$. Solving these systems:

For $O - E = 11$:
$$2O = 42 \Rightarrow O = 21, E = 10$$

For $O - E = -11$:
$$2O = 20 \Rightarrow O = 10, E = 21$$

### The Combinatorial Hunt

Now, we identify the sets. In Case $1$, we need $3$ digits for the even positions that sum to $10$.

By systematically checking combinations, we find: $\{1, 2, 7\}$, $\{2, 3, 5\}$, and $\{1, 4, 5\}$. These are $3$ distinct sets.

In Case $2$, we need $3$ digits that sum to $21$. The only combination is $\{5, 7, 9\}$.

### The Final Symphony

Finally, we calculate the arrangements. For any chosen set of $3$ digits in the even positions, there are $3!$ ways to arrange them.

The remaining $4$ digits for the odd positions can be arranged in $4!$ ways. Thus, for each set, the number of arrangements is:

$$3! 	imes 4! = 6 	imes 24 = 144$$

Since we have $3$ sets in Case $1$, we have $3 	imes 144 = 432$ numbers. In Case $2$, we have $1 	imes 144 = 144$ numbers.

Adding these together, the total number of arrangements is:

$$432 + 144 = 576$$

We have conquered the problem. The final answer is **576**.

Animated Solution for Mathematics - Permutations and Combinations: The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is ____.

Enter Numerical Value:

Visualized Solution

Setting up the $7$ -Digit Number

Sum of Available Digits

Divisibility Rule of $11$

Equations for $O$ and $E$

Finding Possible Values for $O$ and $E$

Case 1: $E = 10$

Arrangements for Case 1

Case 2: $E = 21$

Arrangements for Case 2

Final Calculation

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Alternating Balance

The Algebraic Constraint

The Combinatorial Hunt

The Final Symphony

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