Math 221 Final Exam Review

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1 Math 221 Final Exam Review Preliminary comment: Some of these problems a formulated using language and structures from graph theory. However they are generally self contained; no theorems from graph theory are needed. They are essentially applications of the counting methods that we have discussed all quarter. The necessary graph theoretic definitions are provided here. 1. Define a simple labelled graph on V to be a pair (V, E where V is a non-empty set, and E is a (possibly empty set of 2-element subsets of [V ]. The elements of E are called edges, and the elements of V are called vertices (singular vertex Thus E {S P owerset([n] : S = 2}. (When we did this problem in class, I drew pictures for small values of n. For most people it helps to visualize the construction of the graphs in stages. (a What is the maximum number of edges that a simple labelled graph on [n] can contain? 2 (b How many simple labelled graph on [n] contain exactly edges? ( ( n 2 (c How many simple labelled graphs on [n] are there? 2 (n 2 (d How many simple labelled graph on [n] have an even number of edges? (n. (Either by a bijection or by the binomial theorem. (e How many simple labelled graph on [n] have the property that {1, 2} is an edge, but {1, 3} is not? 2 (n 2 2 = 1 4 2(n 2. (There are only 2 2 decisions to mae, instead of If v is a vertex in simple labelled graph, define degree(v to be the number of edges that contain v. Let n 2017.

2 (a How simple labelled graphs on [n] have the property that vertex 221 has degree? (When we did this problem in class, I drew pictures showing how the graph is built up by first choosing the neighbors of vertex 221, then noting that the rest of the graph is just a simple labelled graph on n 1 vertices. (b How many simple labelled graphs have the property that no edge has both of its vertices odd? In other words, v, w E v w 0 mod 2. (You many assume that n is even. (This problem was done in class Let n = 2m; there are m vertices with even labels, and m vertices with odd labels. Then there are m 2 possible edges with one even vertex and one one odd vertex. There are also ( m 2 possible edges with both vertices odd. When forming a graph, for each of the m 2 + ( m 2 possible edges, we have two choices: include the edge or not. Hence the answer is 2 m2 +( m 2. (c How many simple labelled graphs on [n] have the property that no vertex has degree 0? We did this tough PIE problem in class. Let Ω be the set of all simple labelled graphs on [n]. Let A i = {G : vertex i is isolated}. n The problem is to determine the cardinality of A i. If S is any set of vertices, then the number of graphs in which all the vertices in S have degree 0 is 2 (n 2. Hence, by inclusion exclusion, n A i = Ω A i A i A j +... i=1 i=1 i<j ( ( = 2 (n 2 n 2 (n 1 2 n + 2 (n ( n = 2 (n j 2. j j=0 i=1 2

3 I don t now a way to simplify this expression. However you might be interested to now that, if n is large, the first term in the sum is much larger than all the rest combined. Most graphs don t have any vertices of degree Multigraphs are a generalization of graphs in which more than one edge is permitted between two vertices. More formally, a simple labelled multigraph on V is a pair (V, E where V is a non-empty set, and E is a multiset of 2-elements subsets of V. (a How many multigraphs on [n] have exactly edges? For each edge {u, v}, let X {u,v} be the number of edges between u and v. Then the problem is to determine the number of solutions to the equation X {u,v} = {u,v} The number of terms in the sum is 2. So we can apply the counting table directly to get the answer: ( (b How many multigraphs on [n] have edges with vertex 1 having degree d? Mae sure you understand problem 2a before doing this one. This one is more complicated because repeated edges are permitted. First lets count the number of ways to choose that edges that contain vertex 1. For each v > 1, let Y v be the number of copies of the edge {1, v}. Then each Y v is non-negative and n v=2 Y v = d From the counting table, the number of ways to choose the d neighbors of vertex 1 is ( d+(n 1 1 n 2. The rest of the graph (the d edges that do not contain vertex 1 form a simple labelled multigraph on [n 1] with d edges. By the same argument as in part (a, the number of ways to choose this part of the graph is ( ( d Putting the pieces

4 together, we get the final answer: Whew! ( d + (n 1 1 n 2 ( ( d A directed graph on V is pair (V, E where V is a non-empty set and E V V. These differ in two ways from simple labelled graphs. First the edges are ordered pairs of vertices. So (1, 3 and (3, 1 are considered different edges. We also allow loops : edges (v, w where v = w Warning: This is not quite the same as the problem we discussed in class because we are allowed to have both (u, v and (v, w. (In class I only allowed one of the two. So it is very similar, but the answers are different, (a How many directed graphs on [n] are there? 2 n2. (b How many directed graphs on [n] have edges (including loops? 2. (c How many directed graphs on [n] have the property that vertex n is not in any edge? 2 (n For each of the following, determine the number of wals with the stated properties, where each step in the wal is either one to the right or one up. (a The wal begins at 0, 0 and ends at (n, n ( 2n n (b The wal begins at 0, 0, ends at (2n, 2n, and passes through (n, n. ( 2n n 2 4

5 (c The wal begins at 0, 0, ends at (2n, 2n, and does not through (n, n. ( 4n ( 2n 2n 2 n 6. Find a recurrence for the number of ways to tile a 1 n board with 1 1 squares and 1 2 dominos, given that the squares come in x colors and the dominoes come in y colors. a 1 = x, a 2 = x 2 + y; for n > 2, a n = xa n 1 + ya n Explain why there cannot be integers x and y such that 4x 2 + 2y 10 16y 3 x = 5. Left side is even, right side is odd. 8. What is the remainder when is divided by 7? No electronic devices permitted. Hint: 2 5 mod 7. zero ( = ( = Suppose F (x = n=0 a n x n = 5 1 4x x x (a Find an explicit formula for a n a n = 5 4 n n + ( 1 n (b Find a recurrence relation that the numbers a n satisfy. a 0 = 8, a 1 = 25, a 2 = 99, and for all n 3, a n = 6a n 1 5a n 2 12a n You are dealt 5 cards from a standard dec of cards. (a How many possibilities are there for the 5 card hand that you are dealt? (

6 (b How many of these possibilities include exactly three cards with the same single digit, but no other cards with numbers. (So, for example, we count the hand {A, 3, 3, 3, K }, but not the hand {A, 10, 10, 10, K }, and not {A, 3, 3, 3, 5 }. The number of cards that are not numbers is 4 4 = 16, so we get 8 (4 16 3( Suppose 5 decs of cards are shuffled together, and you are dealt a five card hand. How many possibilities are there for the multiset of 5 cards that you are dealt? Let X i be the number of copies of card i. The number of solutions to 52 X i = 5 is ( ( = 56 i= Let A be the set of all solutions to the equation X 1 + X 2 + X 3 = 100, where each X i is a non-negative integer. Let B be the set of all solutions to the equation Y 1 + Y 2 + Y 3 + Y 4 = 200 where each Y i is an integer and Y 1 1, Y 2 2, Y 3 3 and Y 4 = 94. (a Carefully define a bijection F : A B. Also define its inverse G : B A. F ((X 1, X 2, X 3 = (X 1 + 1, X 2 + 2, X 3 + 3, 94. G(Y 1, Y 2, Y 3, Y 4 = (Y 1 1, Y 2 1, Y 3 1 (b What is F ((3, 0, 97? (4, 2, 100, 94 (c What is G((4, 2, 100, 94? (3, 0, You begin at the center of a chess board and wal until you reach one of the four corners of the board. The rules are that each step involves traversing one side of a square, beginning wherever you are (So, unless you are at the edge of the board, there are four choices of which direction to tae. Let a n be the number of wals of length n. Show that a n is even for all positive integers n. Is a n 0 mod 4? 6

7 Let b NE n, b SE n, b SW n, b NE n repectively be the number of wals of length n that end at the upper right, lower right, lower left, and upper left corners of the board. Thus a n = b NE n + b SE n + b SW n + b NE n. If we draw the path traced by a NE wal, and rotate the board 90 degrees clocwise, the we get the path of a uniquely determined SE wal. Hence b NE = b SE. By a similar argument (rotate 180 or 270 degrees. b NE = b SW = b NW. Thus a n = 4b SE, a multiple of 4. Every multiple of 4 is even, so we have both a n 0 mod 4 and a n 0 mod Let a n be the number of simple labelled graphs on [n] in which each vertex has degree one. (a What is a 2017? Hint: 2017 is odd. a 2017 = 0 (b Find a recurrence relation that the numbers a n satisfy. a 1 = 0, a 2 = 1, and for all n 3, a n = (n 1a n 2. (c Find an exact formula for a n. m (2 1,if n = 2m a n = =1 0,if n is odd (d Prove that a 2m 2 m m! for all positive integers m. Use part (c and the inequality 2 1 < 2 : a 2m = m (2 1 =1 m (2 = 2 m =1 m =1 = 2 m m! 15. Let a n be the number of strings of length n over the alphabet {A, B, C, D} in which A never appears twice consecutively. 7

8 (a Find a recurrence relation that the numbers a n satisfy. a 1 = 4, a 2 = 15, and for all n 3, a n = 3a }{{ n 1 } + 3a }{{ n 2 } last not A ends in A (b Find an explicit formula for a n This one is unreasonably messy for hand computations. Sorry. 16. The binomial theorem asserts that, when (x + y n is expanded, the coefficient of x y n is ( n (a Prove it using induction First remember the identify ( ( m m 1 = 1 ( m 1 + that was proved in class using a bijective argument. Since ( ( 1 0 = 1 1 = 1, the base case (x + y 1 = 1 x y 1 is clear. =0 Assume, as the inductive hypothesis, that (x + y n = =0 (1 ( n x y n. (2 8

9 Then (x + y n+1 = (x + y(x + y n = x(x + y n + y(1 + x n ( n }{{} = x x j y n + y j I.H j=0 =0 ( n = x j+1 y n+1 (j+1 + j j=0 =0 n+1 ( n = x y n =1 =0 n+1 ( n n+1 = x y n =0 =0 x y n x y n+1 x y n+1 x y n The last step was a little tricy because of the limits of summation. In the first sum, we haven t really changed anything by adding and extra = 0 term because 1 = 0 Similarly the second ( sum is unchanged when we add an extra = n + 1 term because n n+1 = 0. The reason for fiddling with the indices is so that we can combine the two sums into one: n+1 (( n (x + y n+1 = + =0 ( n x y n+1 (3 1 Now we can simply apply the identity (1 with m = n + 1: n+1 ( n + 1 (x + y n+1 = x y n+1 (b Use it to show that 0 2 =0 =0 = 0 With x = 1, y = 1, we get ( n 0 n = ( 1 = even 2+1 ( n 1 + odd ( n ( 1 9

10 (c Use a bijection to prove the same fact. Define ϕ : P owerset([n] P owerset([n] by { S {n} if n S ϕ(s = S {n} if n S 17. If (x+y +z +w m is expanded, and lie terms are combined, how many nonzero terms will there be? Assuming the variables commute, this is just the number of solutions to X 1 + X 2 + X 3 + X 4 = m with each X i 0, namely ( ( 4+m = m