SOLUTION: A permutation of the numbers (1,2,3,\dots,n) is a rearrangement of the numbers in which each number appears exactly once. For example, (2,5,1,4,3) is a permutation of (1,2,3,4,5).

Question 1210200: A permutation of the numbers (1,2,3,\dots,n) is a rearrangement of the numbers in which each number appears exactly once. For example, (2,5,1,4,3) is a permutation of (1,2,3,4,5).

Find the number of permutations (x_1, x_2, \dots, x_8) on (1, 2, 3, \dots, 8), such that (x_i, x_{i + 1}), 1 \le i \le 7, is never equal to (1,2), (3,4), (5,6), or (7,8). (However, (x_i, x_{i + 1}) can be equal to (2,1), (4,3), (6,5), or (8,7).)
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem using the Principle of Inclusion-Exclusion.
**1. Total Number of Permutations**
* The total number of permutations of (1, 2, 3, ..., 8) is 8! = 40320.
**2. Define the Sets**
Let:
* A₁ be the set of permutations where (1, 2) appears.
* A₂ be the set of permutations where (3, 4) appears.
* A₃ be the set of permutations where (5, 6) appears.
* A₄ be the set of permutations where (7, 8) appears.
We want to find |S \setminus (A₁ ∪ A₂ ∪ A₃ ∪ A₄)|, where S is the set of all permutations.
**3. Apply Inclusion-Exclusion Principle**
* |S| = 8!
* |Aᵢ| = 7! (treat (1, 2), (3, 4), (5, 6), (7, 8) as single units)
* |Aᵢ ∩ Aⱼ| = 6! (treat two pairs as single units)
* |Aᵢ ∩ Aⱼ ∩ A_k| = 5! (treat three pairs as single units)
* |A₁ ∩ A₂ ∩ A₃ ∩ A₄| = 4! (treat all four pairs as single units)
By the Principle of Inclusion-Exclusion:
|A₁ ∪ A₂ ∪ A₃ ∪ A₄| = Σ |Aᵢ| - Σ |Aᵢ ∩ Aⱼ| + Σ |Aᵢ ∩ Aⱼ ∩ A_k| - |A₁ ∩ A₂ ∩ A₃ ∩ A₄|
|A₁ ∪ A₂ ∪ A₃ ∪ A₄| = ⁴C₁ * 7! - ⁴C₂ * 6! + ⁴C₃ * 5! - ⁴C₄ * 4!
**4. Calculate the Values**
* 7! = 5040
* 6! = 720
* 5! = 120
* 4! = 24
|A₁ ∪ A₂ ∪ A₃ ∪ A₄| = 4 * 5040 - 6 * 720 + 4 * 120 - 1 * 24
|A₁ ∪ A₂ ∪ A₃ ∪ A₄| = 20160 - 4320 + 480 - 24
|A₁ ∪ A₂ ∪ A₃ ∪ A₄| = 16320 + 456
|A₁ ∪ A₂ ∪ A₃ ∪ A₄| = 16296
**5. Find the Desired Number of Permutations**
* Number of permutations = |S| - |A₁ ∪ A₂ ∪ A₃ ∪ A₄|
* Number of permutations = 8! - 16296
* Number of permutations = 40320 - 16296
* Number of permutations = 24024
**Therefore, there are 24024 permutations that satisfy the given conditions.**

Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.

The problem is posed/worded/ INCORRECTLY.

To be correct, it should talk consistently not about permutations, but about the sequences of integer numbers.

Since it is worded in wrong way, I even will not discuss anything here.

It looks like the problem's creator does not understand the difference.

In such case, it is useless to make any explanations.

Simply, the problem's creator did not achieve the level yet,
when such discussing can be productive.

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