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Final Exam with Solution - Discrete Mathematics | MATH 55, Exams of Discrete Mathematics

University of California - BerkeleyDiscrete Mathematics
Prof. L. K. Williams

Material Type: Exam; Professor: Williams; Class: Discrete Mathematics; Subject: Mathematics; University: University of California - Berkeley; Term: Spring 2009;

Typology: Exams

2010/2011

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Math 55, Final Exam
Wednesday, May 20, 8:00am–11:00am
This exam is closed book. Do not use any books, notes or
electronic devices. Please write in a blue note book, with your
name, the name of your TA and your section time on the front.
Each of the 10 problems is worth 10 points, for a total of 100
points. Answers without justification will receive no credit.
(1) (a) How many nonisomorphic unrooted trees are there
with 4 vertices ?
(b) How many nonisomorphic rooted trees are there with
4 vertices ?
(2) How many ways are there to distribute five balls into three
boxes if each box must have at least one ball in it and
(a) the balls are labeled but the boxes are unlabeled?
(b) the balls are unlabeled but the boxes are labeled?
(c) both the balls and the boxes are unlabled?
(3) Give a recursive definition of the set of bit strings that are
palindromes.
(4) Consider the divisibility poset {1,2,3,12,18,36},|.
(a) Draw the Hasse diagram of this poset.
(b) Determine whether this poset is a lattice.
(5) Determine the number of paths in the xy plane starting
at the origin (0,0) and ending at the point (m, n). Here,
each path is made up of a series of steps, where each step
pf3
pf4

Partial preview of the text

Download Final Exam with Solution - Discrete Mathematics | MATH 55 and more Exams Discrete Mathematics in PDF only on Docsity!

Math 55, Final Exam

Wednesday, May 20, 8:00am–11:00am

This exam is closed book. Do not use any books, notes or

electronic devices. Please write in a blue note book, with your

name, the name of your TA and your section time on the front.

Each of the 10 problems is worth 10 points, for a total of 100

points. Answers without justification will receive no credit.

(1) (a) How many nonisomorphic unrooted trees are there

with 4 vertices?

(b) How many nonisomorphic rooted trees are there with

4 vertices?

(2) How many ways are there to distribute five balls into three

boxes if each box must have at least one ball in it and

(a) the balls are labeled but the boxes are unlabeled?

(b) the balls are unlabeled but the boxes are labeled?

(c) both the balls and the boxes are unlabled?

(3) Give a recursive definition of the set of bit strings that are

palindromes.

(4) Consider the divisibility poset

(a) Draw the Hasse diagram of this poset.

(b) Determine whether this poset is a lattice.

(5) Determine the number of paths in the xy plane starting

at the origin (0, 0) and ending at the point (m, n). Here,

each path is made up of a series of steps, where each step

is a move one unit to the right or a move one unit upward.

(6) A new employee checks the hats of six people at the opera,

forgetting to put claim check numbers on them. When

people come back for their hats, the checker returns hats

chosen at random from the remaining hats.

(a) What is the probability that no one receives his own hat?

(b) What is the expected number of hats that are returned

correctly?

(7) Find an integer x such that

x ≡ 0 (mod 2), x ≡ 3 (mod 7) and x ≡ 6 (mod 11).

(8) Let G 1 and H 1 be two isomorphic graphs, and let G 2 and

H 2 be two isomorphic graphs. Prove or disprove that the

graphs G 1 ∪ G 2 and H 1 ∪ H 2 are isomorphic.

(9) There are three 6-sided dice on a table. Two are regular

fair dice, and the third is a loaded die that comes up 6 half

of the time. You grab a die at random, try it, and roll a

6. What is the probability that you have the loaded die?

(10) How many relations are there on the set A = { 1 , 2 , 3 } that

(a) are symmetric?

(b) are reflexive and symmetric?

(c) are neither reflexive nor irreflexive?

strings of length m + n having m zeros. That number is the binomial coefficient

(m+n m

. [# 33 in §5.4, page 369]

  1. (a) This is derived in Table 2 in §7.6 on page 512:

(b) The expected number of hats that are returned correctly equals 1 , regardless of how many people check their hats at the opera. This is derived from linearity of expectation in Example 6 of §6.4 on page 430.

  1. From the identifies (−38) · 2 + 1 · 77 = 1, (−3)7 + 1 · 22 = 1 and (−5) · 11 + 4 · 14 = 1, we see that 77 ≡ 1 (mod 2), 22 ≡ 1 (mod 7), and 4 is the inverse to 14 modulo 11. By the method used in the proof of the Chinese Remainder Theorem, x = 0 · 77 + 3 · 22 + 6 · 4 · 14 = 402 is one solution, and all solutions have the form x = 402 + 154n for some integer n. Taking n = −2 gives the smallest positive solution x = 94.
  2. This statement is false. We can disprove it by giving a counterexam- ple. The following four graphs all have vertex set V = { 1 , 2 , 3 , 4 }. Take G 1 = H 1 = (V, {{ 1 , 2 }}), G 2 = (V, {{ 2 , 3 })), and H 2 = (V, {{ 3 , 4 }}). All four graphs are isomorphic. However, G 1 ∪G 2 = (V, {{ 1 , 2 }, { 2 , 3 }}) is connected while H 1 ∪ H 2 = (V, {{ 1 , 2 }, { 3 , 4 }}) is disconnected, so they are not isomorphic. [# 20 in §9.Supp on page 679]
  3. We apply Bayes’ Theorem. Let E be the event that a 6 is observed, and let Di be the event that the ith dice is chosen, for i = 1, 2 , 3. Each die is equally likely, p(D 1 ) = p(D 2 ) = p(D 3 ) = 1/3. We also know that p(E|D 1 ) = p(E|D 2 ) = 1/6 while p(E|D 3 ) = 1/2. Bayes’ Theorem says

p(D 3 |E) =

p(E|D 3 )p(D 3 ) p(E|D 1 )p(D 1 ) + p(E|D 2 )p(D 2 ) + p(E|D 3 )p(D 3 )

So, the answer is p(E|D 1 ) = (^) (1/6)(1/3)+(1(1//2)(16)(1//3)3)+(1/2)(1/3) = 3 / 5.

  1. For an n-element set this appears in # 45 of §8.1 on page 529. We get

(a) 2(

4 2 ) = 2^6 = 64 (b) 2(

3 2 ) = 2^3 = 8 (c) 2^9 − 2 · 23 ·^2 = 512 − 128 = 384.