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Let n be a positive integer. A permutation a of the symmetric group of permutations of is called a derangement if for each . Suppose that x and y are two arbitrary permutations of . We say that a permutation a is a double derangement with respect to x and y if and for each . In this paper , we give an explicit for mula for , the number of double derangements with respect to x and y. Let and let and be two subsets of with and . Suppose that denotes the number of derangements x such that . As the main result, we show that if and z is a permutation such that for and for , then where .

Let n be a positive integer. A derangement is a permutation of the symmetric group

The number of derangements also satisfies the relation

exclusion principle that

These facts and some other results concerning derangements can be found in [

exactly k fixed points. The number of such permutations is denoted by

concerning the permutations of

Let e be the identity element of the symmetric group

In the following, we assume that n is a positive integer and the identity permutation of the symmetric group

Definition 1. Suppose that x and y are two arbitrary permutations of

Proposition 1. Let

Proof. Let

Case 1.

Case 2.

These considerations show that

We can therefore assume that

For a derangement x satisfying

If the first case occurs then we have to evaluate the number of derangements of the set

If the second case occurs then we have to evaluate the number of derangements of the set

We now use induction on k to show that

For

Now let the result be true for

Corollary 1. Let k be a positive integer. Then

Proof. Let

The following

The following lemma can be easily proved.

Lemma 1. Let x and y be two arbitrary permutations and

Theorem 2. Let

n\k | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

1 | 0 | 0 | 0 | 0 | 0 |

2 | 1 | 0 | 0 | 0 | 0 |

3 | 1 | 0 | 0 | 0 | 0 |

4 | 3 | 2 | 0 | 0 | 0 |

5 | 11 | 4 | 0 | 0 | 0 |

6 | 53 | 14 | 6 | 0 | 0 |

7 | 309 | 64 | 18 | 0 | 0 |

8 | 2119 | 362 | 78 | 24 | 0 |

9 | 16,687 | 2428 | 426 | 96 | 0 |

10 | 148,329 | 18,806 | 2790 | 504 | 120 |

where

Proof. Let

and

where

Our ultimate goal is to find an explicit formula for evaluating

Lemma 2. Let

Moreover,

Proof. We can restate the problem as follows: We want to put k ones and

the

the i-th block be

Now suppose that we write

Lemma 3. Let

Moreover,

Proof. Similar to the above argument, we want to put k ones and

Case 1. There is no block of ones before the first zero and after the last zero. In this case we put

and choose

ways. Let the number of ones in the i-th block be

solutions for the latter equation is

Case 2. There is no block of ones before the first zero but there is a block after the last zero. In this case we put

ones in

Case 3. There is a block of ones before the first zero but there is no block after the last zero. This is similar to the above case.

Case 4. There is a block of ones before the first zero and a block of ones after the last zero. In this case we must have

places for putting

These considerations prove that

A straightforward computation gives the result.

The following

k\l | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|---|

0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

2 | 35 | 10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

3 | 50 | 60 | 10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

4 | 25 | 100 | 75 | 10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

5 | 2 | 40 | 120 | 80 | 10 | 0 | 0 | 0 | 0 | 0 | 0 |

6 | 0 | 0 | 25 | 100 | 75 | 10 | 0 | 0 | 0 | 0 | 0 |

7 | 0 | 0 | 0 | 0 | 50 | 60 | 10 | 0 | 0 | 0 | 0 |

8 | 0 | 0 | 0 | 0 | 0 | 0 | 35 | 10 | 0 | 0 | 0 |

9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 10 | 0 | 0 |

10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |

Theorem 3. Let c be be a cycle of length

Proof. Let

Using the notations of Theorem 2,

Example 1. We evaluate

and

Applying Theorem 3 with

and

The above example shows that how can we evaluate

Pooya Daneshmand,Kamyar Mirzavaziri,Madjid Mirzavaziri, (2016) Double Derangement Permutations. Open Journal of Discrete Mathematics,06,99-104. doi: 10.4236/ojdm.2016.62010