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Problem 1
Consider 101 sets with the following properties: (i) each set has 1000 elements; (ii) the intersection of any two distinct sets has 20 elements; (iii) the intersection of any three pairwise distinct sets has 6 elements; (iv) the intersection of four or more pairwise distinct sets is empty. How many elements are there in the union of the union of these 101 sets? Give a numerical answer. (The calculations are not involved and the answer turns out to have a surprisingly simple form. (You’ll know it when you see it.) Brute force attempts won’t work here, so don’t even try!)
Solution: Denoting the sets by A 1 ,... A 101 , we are given that |Ai| = 1000, |Ai∩Aj | = 20, |Ai ∩ Aj ∩ Ak| = 6, and |Ai ∩ Aj ∩ Ak ∩ Al| = 0 for any pairwise distinct i, j, k, l. We need to find. | ∪^101 i=1 Ai|. By inclusion/exclusion, this is ∣ ∣∣ ∣ ∣
i=
Ai
∑^101
i=
|Ai| −
1 ≤i<j≤ 101
|Ai ∩ Aj | +
1 ≤i<j<k≤ 101
|Ai ∩ Aj ∩ Ak|
Problem 2
- Write down an exact formula for Dn, the number of derangements of n objects. Solution: Dn = n!
∑^ n
k=
(−1)k k!
- Write down an approximate value for Dn (involving a famous constant) when n is large. Solution: Dn ≈ n!e−^1.
- How many ways are there to list the numbers 1, 2 ,... , 20 in some order such that exactly 5 of these numbers are in their “natural” positions? For example, 7 is in its natural position if it is the 7-th number in the list. Your answer can be left in raw form, and may involve factorials, binomial coefficients, as well as the function Dn, but no summations. Solution: There are
5
ways to pick the numbers that are to remain in their natural positions, and D 15 ways to derange the remaining 15 numbers, giving a
total of
D 15 ways for an arrangement of the requested type.
Problem 3
Consider the relation R on the set { 1 , 2 , 3 , 4 } defined by the matrix
(That is, (i, j) ∈ R if the ij-th entry in this matrix is 1, and (i, j) 6 ∈ R if the ij-th entry is 0.) Is this relation an equivalence relation on the set { 1 , 2 , 3 , 4 }? If it is, explain briefly why it has each of the relevant properties. If not, state which of the properties of an equivalence relation it fails, and for each of these properties give a concrete example for which the property does not hold.
Solution: The properties required of an equivalence relation are (i) reflexivity, (ii) symmetry, and (iii) transitivity. For each of these we check whether it holds:
(i) The relation is not reflexive since the diagonal entries are all 0; for example, (1, 1) 6 ∈ R.
(ii) The relation is symmetric since the matrix is symmetric.
(iii) The relation is not transitive, since, for example, (2, 1) ∈ R, (1, 4) ∈ R, but (2, 4) 6 ∈ R. (Examples such as these can be easily discovered by drawing the graph representing the relation.)
Problem 4
Consider the graph K 10 , the complete graph with 10 vertices.
- How many edges does this graph have? (Hint: Don’t try to draw the graph and count!) Solution: There is one edge corresponding to each unordered pair of vertices, so the number is
2
= 45. (Alternatively, one can argue via the handshake theorem: There are 10 vertices, each having degree 9, so the sum of all degrees is 90. By the handshake theorem, this is twice the number of edges, so there are 90/2 = 45 edges.)
- Does this graph have an Euler circuit? Explain! Solution: No. An Euler circuit exists only if the graph is connected and each vertex has even degree. In the given graph, each vertex has degree 9 (as there are 9 edges connecting this vertex to the other 9 vertices), so, in fact, none of the vertices has even degree.
Problem 5
Consider the following pairs of graphs. In each case, determine whether the graphs are isomorphic. If they are, give a one-to-one matchup of vertices. If they are not, explain clearly why they cannot be isomorphic.
