Double Derangement Permutations

Let n be a positive integer. A permutation a of the symmetric group n S of permutations of [ ] { } n n 1, 2, , =  is called a derangement if ( ) a i i ≠ for each [ ] i n ∈ . Suppose that x and y are two arbitrary permutations of n S . We say that a permutation a is a double derangement with respect to x and y if ( ) ( ) a i x i ≠ and ( ) ( ) a i y i ≠ for each [ ] i n ∈ . In this paper, we give an explicit formula for ( ) n D x y , , the number of double derangements with respect to x and y. Let k n 0 ≤ ≤ and let { } k i i 1 , ,  and { } k a a 1 , ,  be two subsets of [ ] n with j j i a ≠ and { } { } k k i i a a 1 1 , , , , =     . Suppose that ( ) n k , , ∆  denotes the number of derangements x such that ( ) j j x i a = . As the main result, we show that if m n 0 ≤ ≤ and z is a permutation such that ( ) z i i ≠ for i m ≤ and ( ) z i i = for i m > , then ( ) ( ) ( ) ( ) ∑ ∑ k m k n k k i i m D e z n k i i 1 1 01 , 1 , , , , , = ≤ < < ≤ = − ∆


Suppose that ( ) n k , , ∆
 denotes the number of derangements x such that ( ) j j x i a = .As the main result, we show that if m n 0 ≤ ≤ and z is a permutation such that ( ) ( )

D e z n k i i
These facts and some other results concerning derangements can be found in [1].There are also some generalizations of this notion.The problème des rencontres asks how many permutations of the set [ ] n have exactly k fixed points.The number of such permutations is denoted by , n k D and is given by .
Thus, we can say that , . Some probabilistic aspects of this concept and the related notions concerning the permutations of n S is discussed in [2] and [3].Let e be the identity element of the symmetric group n S , which is defined by ( )

The Result
In the following, we assume that n is a positive integer and the identity permutation of the symmetric group n S of permutations of [ ] n is denoted by e.Moreover, for two permutations a and b of n S , the notation a b ⊥ means that ( ) ( ) We also denote the number of elements of a set A by A .Definition 1. Suppose that x and y are two arbitrary permutations of n S .We say that a permutation a is a double derangement with respect to x and y if a x ⊥ and a y ⊥ .The number of double derangements with respect to x and y is denoted by , , , , . Then ( ) , , n k ∆  , the number of derangements x such that ( ) Thus r s a i = for some s r ≠ .Now there are two cases: Case 1.
In this case a derangement x satisfies the condition ( ) for all j t ≠ , where for j s ≠ and s t a a ′ = .This provides a one to one correspondence between the derangements x of [ ] n with ( ) with  elements in their intersections, and the derangements . In this case a derangement x satisfies the condition ( ) x i a ′ = for all j s ≠ .This provides a one to one correspondence between the derangements x of [ ] n with ( ) with  elements in their intersections, and the derangements x′ of [ ] { } \ s n a with ( ) elements in their intersections.
These considerations show that ( ) ( ) . Iterating this argument, we have We can therefore assume that 0 =  .We thus evaluate ( ) , where 2k n ≤ .For 0 k = , we obviously have ( ) For a derangement x satisfying ( ) j j x i a = there are two cases: ( ) x a i = or ( ) x a i ≠ .
If the first case occurs then we have to evaluate the number of derangements of the set [ ] { } with 0 elements in their intersections.The number is equal to If the second case occurs then we have to evaluate the number of derangements of the set with 0 elements in their intersections.The number is equal to ,1, 0 1, 0, 0 2, 0, 0 . 1 Now let the result be true for 1 k − .We can write ) .
Let k be a positive integer.Then x i a = if and only if x′ defined by ( ) ( ) The number of such permutations x′ is !k .
The following Table 1 gives some small values of ( ) The following lemma can be easily proved.Lemma 1.Let x and y be two arbitrary permutations and 0 m ≥ be the number of i's for which ( ) ( ) Then there is a permutation z such that ( ) and ( ) and let z be a permutation such that ( ) and ( ) where ( , , , , , , Let i E be the set of all derangements x for which ( ) ( ) , where


. We use the inclusion-exclusion principle to determine , , , , , . This implies the result.
Our ultimate goal is to find an explicit formula for evaluating for other values of , , n k  .Proof.We can restate the problem as follows: We want to put k ones and n k − zeros in a row in such a way that there are exactly  appearance of one-one.To do this we put n k − zeros and choose k −  places of . The number of solutions for the latter equation is . Now suppose that we write 1, 2, , n  around a circle.We thus assume that 1 is after n and so ,1 n is also assumed to be consecutive.Under this assumption we have the following result.
Lemma 3. Let ( ) for other values of , , n k  .
Proof.Similar to the above argument, we want to put k ones and n k − zeros around a circle in such a way that there are exactly  appearances of one-one.At first, we put them in a row.There are four cases: Case 1.There is no block of ones before the first zero and after the last zero.In this case we put n k − zeros and choose k −  places of the Let the number of ones in the i-th block be 1 i r ≥ .We then must have 1 . The number of solutions for the latter equation is There is no block of ones before the first zero but there is a block after the last zero.In this case we put n k − zeros and choose ways.Let the number of ones in the i-th block be 1 i r ≥ .We then must have . The number of solutions for the latter equation is 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 1 −  appearance of one-one in the row format, since we want to achieve  appearance of one-one in the circular format.Thus we put n k − zeros and choose ( ) ways.Let the number of ones in the i-th block be 1 i r ≥ .We then must have ( ) . The number of solutions for the latter equation is 1 1 These considerations prove that ( ) The following Table 2 gives some small values of ( )  Using the notations of Theorem 2, ( ) Applying Theorem 3 with 3 m = we have x and y are two arbitrary permutations of n S .We say that a permutation a is a double derangement with respect to x and y if ( ) ( ) Let n be a positive integer.A derangement is a permutation of the symmetric group n S of permutations of[ ] { } 1, 2, , n n =  such that none of the elements appear in their original position.The number of derangements of n S is denoted by n D or n¡.A simple recursive argument shows that It can be proved by the inclusion- exclusion principle that n D is explicitly determined by ( ) for an arbitrary cycle c.Prior to that we need to state two elementary enumerative problems concerning subsets A of the set [ ] n with k elements and exactly  consecutive members.exactly  solutions for r and s in A. If 0

Theorem 3 .
Let c be be a cycle of length m n m has exactly  solutions for the equation 1 r s ≡ + (mod n) for , r s in A. Thus the number of { }  .Applying Theorem 2, we have the result.double derangements x with respect to e and 5 c are . In the present paper, we extend the concept of a derangement to a double derangement with respect to two fixed elements x and y of n S .
n D is the number of a with a e ⊥ .If c is any fixed element of n S then the number of