On finding a particular class of combinatorial identities

A modification of the method of "counting in two ways" ( [1], p.2) is to obtain a general formula about the number of elements of a set, after that to consider some of the subsets of this set, and to find elements of these subsets, on the one hand, by using the general formula, and on the other hand, by using specific properties of the subsets.

We will demonstrate this method by considering and proving the following identities:

(1) (5)

What these identities have in common is the left-hand side that can be written by the expression: It remains to give some of the possible combinatorial interpretations of the expression (6). Boolean (binary, or (0,1)-matrix) is a matrix whose entries are equal to 0 or 1. We will show that R(n × k, p) gives the number of all n × k (composed of n rows and k columns) Boolean matrices, such that in each column they have exactly p ones, and in each row these matrices have at least 1 one. Necessary and sufficient condition for existence of such matrices is 1 ≤ p ≤ n ≤ kp, as in the special case when n = kp, in each row we have exactly 1 one, and when n > kp, such matrices do not exist, that is, their number is equal to zero (see Proposition 3). Indeed, it is easy to see that the number r(n × k, p) of all n × k Boolean matrices, that have exactly p ones in each column, is equal to

From the set of all n×k Boolean matrices, that have exactly p ones in each column, we have to remove matrices that have at least one row of zeroes, and to determine the number of the remaining matrices. We will do this by using the principle of inclusion and exclusion. Recall this known principle in the following form:

Inclusion and exclusion principle: Let M be a finite set and let P = {p 1 , p 2 , . . . , p m } be the set of properties that can be possessed by the elements of M . Denote by N (p i 1 , p i 2 , . . . , p is ) the number of elements of M that possess the properties p i 1 , p i 2 , . . . , p is . Then the number N (∅) of elements of M that do not possess any of the properties of P is given by the formula:

In particular, in our problem let M be the set of all n×k Boolean matrices that have in each column exactly p (1 ≤ p ≤ n) ones, and the property p i is possessed by these matrices belonging to the set M whose i-th row, 1 ≤ i ≤ n, consists of zeroes. Since M cannot contain matrices with more than np rows of zeroes, then N (p i 1 , p i 2 , . . . , p is ) = 0 when s > np. Then obviously |M | = r(n × k, p), and for each set of properties {p i 1 , p i 2 , . . . , p is } we have N (p i 1 , p i 2 , . . . , p is ) = r((ns) × k, p), s = 1, 2, . . . , np.

Since s of n properties p 1 , p 2 , . . . , p n can be chosen in n s ways, then

. From (7), applying the inclusion and exclusion principle, we get:

In order to prove the identities (1) -(5), it remains to prove following propositions: Proposition 1. Prove that the number of all quadratic Boolean matrices with n rows and n columns, having in each row and in each column exactly 1 one, is equal to n!.

Proposition 2. Prove that the number of all Boolean matrices with 2p rows and two columns, having in each row 1 one and 1 zero and the same number of ones in each column, is equal to

Proposition 3. Prove that in a Boolean matrix with n rows and k columns, such that in each column there is exactly p ones and n > kp, there exists at least one row of zeroes.

Proposition 4. If B is the set of all Boolean matrices with n rows and k columns, such that in each column there is exactly p ones, n = kp and there are no rows of zeroes, prove that |B| = (kp)! (p!) k .

Proposition 5. If E is the set of all Boolean matrices with n rows and k columns, n ≤ k, that have in each column exactly 1 one and there are no rows of zeroes, prove that

Proofs of Proposition 4 and identity (4):

Proof. From n = kp it follows that in each row of a matrix of the set B there is exactly 1 one and -the number of permutations of length n, composed by s different elements, repeated q 1 , q 2 , . . . , q s times, respectively, is given by the expression n! q 1 !q 2 ! . . . q s ! , where n = q 1 + q 2 + . . . + q s ) is equal to

The proof is completed.

From Proposition 4, with the special case when p = 1, (k = 2), we obtain Proposition 1 (Proposition 2).

Proof. Each Boolean matrix of the set E contains k ones -exactly 1 one in the column and at least 1 one in each row. Therefore |E| = R(n × k, 1). If C, C ∈ E is an arbitrary matrix, the ordered k-tuple (c r 1 1 , c r 2 2 , . . . , c r k k ) consists of its nonzero elements, then let t i be the number of ones in the i-th row, where t i ≥ 1, i = 1, 2, . . . , n and t 1 + t 2 + . . . + t n = k. With matrix C, we can uniquely associate the ordered k-tuple (c r 1 1 , c r 2 2 , . . . , c r k k ) or the ordered k-tuple (r 1 , r 2 , . . . , r k ). The number of Boolean matrices of E, that have t i ones in the i-th row, is equal to the permutations (r 1 , r 2 , . . . , r k ), that is, to the number of permutations of the k-tuple (1, . .

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