Let b1, b2, b3, b4 be a 4-element permutation with bi ∈ {1, 2, 3, ..., 100} for 1 ≤ i ≤ 4 and bi ≠ b

Q: Why do we subtract the intersection n(A \cap B)? Can't we just add n(A) and n(B)?

If you simply add n(A) and n(B), you are counting the cases where both conditions are true twice—once in the count for A and once in the count for B. The Principle of Inclusion-Exclusion corrects this by subtracting the overlap exactly once.

Q: How did you determine there are 98 ways to choose the triplet for condition A?

The triplet is (k, k+1, k+2). The smallest value for k is 1 (giving 1, 2, 3) and the largest value for k is 98 (giving 98, 99, 100). Since k can be any integer from 1 to 98, there are exactly 98 possible triplets.

Q: Why is n(A \cap B) equal to 97?

The intersection requires the sequence to be (k, k+1, k+2, k+3). The smallest sequence is (1, 2, 3, 4) where k=1. The largest sequence is (97, 98, 99, 100) where k=97. Thus, there are 97 such sequences.

Question

Let $b_1, b_2, b_3, b_4$ be a 4-element permutation with $b_i \in \{1, 2, 3, \dots, 100\}$ for $1 \le i \le 4$ and $b_i 
e b_j$ for $i 
e j$, such that either $b_1, b_2, b_3$ are consecutive integers or $b_2, b_3, b_4$ are consecutive integers. Then the number of such permutations $b_1, b_2, b_3, b_4$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Elegance of Counting: A Combinatorial Journey

Welcome, fellow traveler on the road to JEE Advanced excellence. Today, we are not just solving a problem; we are peeling back the layers of a beautiful combinatorial puzzle.

We are tasked with finding the number of 4-element permutations $(b_1, b_2, b_3, b_4)$ from the set $S = \{1, 2, 3, \dots, 100\}$ such that either the first three or the last three integers are consecutive. This is a classic "either-or" problem, and it is a perfect stage to showcase the power of the **Principle of Inclusion-Exclusion**.

### Phase 1: Defining the Landscape

Imagine you have a bag containing one hundred distinct tiles, numbered $1$ to $100$. You are asked to draw four tiles and arrange them in a line. We have two specific conditions that make a permutation "valid":

1. **Condition A**: The first three numbers $(b_1, b_2, b_3)$ form a consecutive sequence $(k, k+1, k+2)$.
2. **Condition B**: The last three numbers $(b_2, b_3, b_4)$ form a consecutive sequence $(k, k+1, k+2)$.

We want to find the total number of permutations that satisfy $A$ or $B$. In the language of set theory, we are looking for $n(A \cup B)$.

### Phase 2: The Anatomy of Condition A

Let's isolate Condition A. We need $(b_1, b_2, b_3)$ to be consecutive, which can be represented as $(k, k+1, k+2)$.

The starting integer $k$ can be any value from $1$ to $98$. If $k=1$, we have $(1, 2, 3)$; if $k=98$, we have $(98, 99, 100)$. This gives us exactly $98$ ways to choose the triplet.

We have a fourth position, $b_4$, to fill. Since the problem requires all four numbers to be distinct, $b_4$ cannot be any of the three numbers we just picked. With $100$ total numbers available and $3$ already used, we have $100 - 3 = 97$ choices for $b_4$.

Thus, the total number of permutations satisfying Condition A is:
$$n(A) = 98 	imes 97 = 9506$$

### Phase 3: The Symmetry of Condition B

Now, look at Condition B. The logic is identical. The triplet $(b_2, b_3, b_4)$ must be consecutive.

Again, there are $98$ ways to choose this triplet. And again, $b_1$ must be distinct from the three numbers in the triplet, leaving us with $97$ choices.

So, the total number of permutations satisfying Condition B is:
$$n(B) = 98 	imes 97 = 9506$$

### Phase 4: The Trap of Overcounting

If we simply add $n(A) + n(B)$, we get $9506 + 9506 = 19012$. However, we have counted some permutations twice.

A permutation satisfies *both* Condition A and Condition B if $(b_1, b_2, b_3)$ is consecutive AND $(b_2, b_3, b_4)$ is consecutive. This forces all four numbers to be consecutive: $(k, k+1, k+2, k+3)$.

We must subtract these overlapping cases to avoid overcounting. The starting number $k$ can range from $1$ to $97$.

If $k=1$, we have $(1, 2, 3, 4)$; if $k=97$, we have $(97, 98, 99, 100)$. So, there are exactly $97$ such sequences, meaning:
$$n(A \cap B) = 97$$

### Phase 5: The Final Resolution

Now, we apply the Principle of Inclusion-Exclusion:
$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

Substituting our values:
$$n(A \cup B) = 9506 + 9506 - 97$$
$$n(A \cup B) = 19012 - 97 = 18915$$

And there it is! The total number of valid permutations is **$18915$**. Remember, in combinatorics, the secret is often not just in the counting, but in the careful management of overlaps.

Enter Numerical Value:

Visualized Solution

Understanding the Permutation

Defining the Conditions

Analyzing Condition $A$

Calculating $n (A)$

Analyzing Condition $B$

Calculating $n (B)$

The Overlap: $A \cap B$

Counting $n (A \cap B)$

Applying Inclusion-Exclusion

Final Calculation

Final Answer

The Sigma Insight: Fundamental Principle of Counting

The Elegance of Counting

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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