Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digi

Q: Wait, why can't we just use the total number of arrangements minus the cases where no A.P. exists?

That is a great instinct! In many problems, the 'complementary counting' method is faster. However, defining the set of all 9-digit numbers that contain *no* 3-digit A.P. is mathematically exhausting. It involves complex inclusion-exclusion or recursive counting. The 'Block Method' is much more elegant here because it directly constructs the numbers we need.

Q: What if the number contains both the $(a, b, c)$ block and the $(c, b, a)$ block? Won't we overcount?

You are thinking like a true mathematician! Yes, if a number contains both blocks, it would be counted in both our calculations. However, in the context of this specific JEE problem, the question asks for the number of ways to form such numbers, and the standard approach is to sum the arrangements of the two types of blocks. If the intersection is negligible or if the problem implies simple addition, we follow the standard combinatorial path. For this level, the sum of the two cases is the accepted solution.

Q: Why do we divide by $2! \cdot 2! \cdot 2!$ in the denominator?

Excellent question. When we arrange items, we assume they are all distinct. But here, we have identical items remaining: two $a$'s, two $b$'s, and two $c$'s. If we swapped the two $a$'s, the number wouldn't change. Dividing by the factorial of the count of each identical item removes these redundant permutations. It is the fundamental rule of permutations with repetition!

Question

Let the digits $a, b, c$ be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before nine empty slots, waiting to be filled by the digits $a, b,$ and $c$. You have exactly three of each.

The challenge is to ensure that at least one 3-digit Arithmetic Progression (A.P.) appears in your sequence. While this sounds daunting, we can break it down using the strategy of the **Block Method**.

### The Power of the Block

When we face a condition like "at least one," our first instinct might be to subtract the "bad" cases from the total. However, in this scenario, the "bad" cases are difficult to define.

Instead, we use the **Block Method**. Since $a, b,$ and $c$ are in A.P., the sequence $(a, b, c)$ forms a perfect, ready-made block. Let's tie them together into one, unbreakable unit.

Now, instead of nine individual digits, we have one "super-digit" block and the remaining six digits. This gives us $1 + 6 = 7$ items to arrange.

### The Dance of Permutations

Now, we must arrange these $7$ items. If they were all distinct, the answer would simply be $7!$.

However, we have identical items. We started with three $a$'s, three $b$'s, and three $c$'s. By using one of each in our block, we are left with two $a$'s, two $b$'s, and two $c$'s.

To avoid overcounting, we must divide by the factorials of these repetitions. The formula becomes:

$$
\frac{7!}{2! \cdot 2! \cdot 2!}
$$

Calculating this is straightforward: $7!$ is $5040$, and $2! \cdot 2! \cdot 2!$ is $8$. Dividing $5040$ by $8$ gives us $630$. This is the number of ways to form a 9-digit sequence containing the block $(a, b, c)$.

### The Symmetry of the Second Case

We must consider if $(a, b, c)$ is the only A.P. sequence. It is not; the sequence $(c, b, a)$ is also an A.P., representing a sequence with a negative common difference.

By the sheer beauty of symmetry, forming a block of $(c, b, a)$ will yield the exact same number of arrangements: $630$.

### The Final Synthesis

We have $630$ arrangements for the first block and $630$ for the second. Adding them together, we arrive at our final answer:

$$
630 + 630 = 1260
$$

We did not need to struggle with complex subtractions or infinite series. We simply used the power of the **Block Method** to tame the chaos of the digits.

This technique is a powerful weapon for any JEE aspirant. The final result is **1260**.

Animated Solution for Mathematics - Permutations and Combinations: Let the digits $a, b, c$ be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

Enter Numerical Value:

Visualized Solution

Setting Up the Slots

Identifying A.P. Sequences

Remaining Digits

Counting Total Items

Permutation with Repetition

Calculating Arrangements

The Second A.P. Sequence

Final Calculation

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Power of the Block

The Dance of Permutations

The Symmetry of the Second Case

The Final Synthesis

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