A partition of a set \( A \) is a family of non-empty subsets of \( A \), such that any two distinct subsets in the family are disjoint, and the union of all subsets equals \( A \). We say that a partition of a set of integers \( B \) is [i]separated[/i] if each subset in the partition does [b]not[/b] contain consecutive integers. Prove that, for every positive integer \( n \), the number of partitions of the set \( \{1, 2, \dots, n\} \) is equal to the number of separated partitions of the set \( \{1, 2, \dots, n+1\} \). For example, \( \{\{1,3\}, \{2\}\} \) is a separated partition of the set \( \{1,2,3\} \). On the other hand, \( \{\{1,2\}, \{3\}\} \) is a partition of the same set, but it is not separated since \( \{1,2\} \) contains consecutive integers.

Let $m, n$ be positive integers. Let $S(n,m)$ be the number of sequences of length $n$ and consisting of $0$ and $1$ in which there exists a $0$ in any consecutive $m$ digits. Prove that \[S(2015n,n).S(2015m,m)\ge S(2015n,m).S(2015m,n)\]

A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience draws two cards from two different boxes and announces the sum of numbers on those cards. Given this information, the magician locates the box from which no card has been drawn. How many ways are there to put the cards in the three boxes so that the trick works?

A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience draws two cards from two different boxes and announces the sum of numbers on those cards. Given this information, the magician locates the box from which no card has been drawn. How many ways are there to put the cards in the three boxes so that the trick works?

Let $N$ be the set of positive integers. If $A,B,C \ne \emptyset$, $A \cap B = B \cap C = C \cap A = \emptyset$ and $A \cup B \cup C = N$, we say that $A,B,C$ are partitions of $N$. Prove that there are no partitions of $N, A,B,C$, that satisfy the following: (i) $\forall a \in A, b \in B$, we have $a + b + 1 \in C$ (ii) $\forall b \in B, c \in C$, we have $b + c + 1 \in A$ (iii) $\forall c \in C, a \in A$, we have $c + a + 1 \in B$

In a community of more than six people each member exchanges letters with exactly three other members of the community. Show that the community can be partitioned into two nonempty groups so that each member exchanges letters with at least two members of the group he belongs to.