2!3! (9)( 8x3 ) = 960x 3 ( 720x 3 ) = 1680x 3

Size: px

Start display at page:

Download "2!3! (9)( 8x3 ) = 960x 3 ( 720x 3 ) = 1680x 3"

Conrad Stafford
5 years ago
Views:

1 CSCI 2200 Foundations of Computer Science Spring 2018 Quiz 2 (April 11, 2018) SOLUTIONS 1. [6 POINTS] An NBA team has nine players show up for a game, but only five players can be on the floor to play the game. In how many ways can the five players on the floor be selected? ( ) 9 ANSWER: We choose five players from nine; therefore, the answer is. 5 (See Problem [or F17 Problem 13.19].) 2. [6 POINTS] Which of the graphs below do not exist (i.e., are impossible to construct)? ANSWER: A graph with four vertices and degree sequence [4, 3, 2, 1] is impossible to construct, because a graph with four vertices cannot have a vertex with degree δ 4. The other connected graphs can be constructed. 3. [6 POINTS] What is the coefficient of x 3 in the expansion of (2x + 1) 10 (3 2x) 5? ANSWER: For each of the two parts of (2x + 1) 10 (3 2x) 5, first identify which specific (ith) terms we need. For (2x + 1) 10, we identify i 3 as yielding the x 3 term from (2x) i (1) 10 i. And for (3 2x) 5, we identify i 2 as yielding the x 3 term from (3) i ( 2x) 5 i. Expanding these both with the Binomial Theorem, we have ( ) ( ) 10 5 answer (2x) 3 (1) 7 (3) 2 ( 2x) ! 3!7! (8x3 ) 5! 2!3! (9)( 8x3 ) 960x 3 ( 720x 3 ) 1680x 3 4. [6 POINTS] Given connected graph G with n vertices and n 4, which of the following properties is false? ANSWER: This statement is false: If every vertex of G has degree at least δ 2, there is a cycle of length δ + 1. Consider G with n 4 and degree sequence [2, 2, 2, 2]; this graph is C 4, i.e., it is a cycle of length δ The other statements can be shown to be true. (See Problem ) 1

2 5. [6 POINTS] How many ways can you misspell the word Facebook if you only get the first letter F correct? ANSWER: We need to find all permutations of acebook without double-counting identical words with only the duplicate o characters swapped. Further, we do not want to count acebook since that would yield a correct spelling of Facebook. Therefore, we have answer (Note that this totals 2519.) ( 7 1, 1, 1, 1, 1, 2 ) 1 7! 1!1!1!1!1!2! 1 7! [6 POINTS] Independently generate five random bits b 1 b 2 b 3 b 4 b 5, with P[b i 0] Compute P[ b i 2]. i1 ANSWER: We need to count all permutations of five bits (or binary strings of length 5) whose bits together have a sum of 2. We could list all possibilities; or we could calculate how many ways to choose three 0s and two 1s: count Therefore, our answer is ( 5 3, 2 ) 5! 3!2! [6 POINTS] Flip seven biased coins, each with P[heads] 1 4. You see the first coin land and it lands heads. What is the probability that the seventh coin lands tails? ANSWER: Each coin toss is an independent event. seventh coin lands tails is 1 P[heads] 3 4. Therefore, the probability that the 2

3 8. [6 POINTS] You and a friend take turns rolling an n-sided die, with n N. The first to roll a one wins. Compute your probability to win (as a function of n) if you roll first. ANSWER: In this problem, we have an infinite probability space. Drawing the outcometree and labeling the edge probabilities, we find the probability of rolling a 1 after one roll to be 1 n 1 n. The probability of not rolling a 1 after one roll is n. Therefore, the probability of rolling a 1 after two rolls is n 1 n 1 n n 1. n 2 Similarly, the probability of rolling a 1 after three rolls is n 1 n 2 Further, the probability of rolling a 1 after four rolls is (n 1)2 n 3 1 n (n 1)2 n 3. 1 n (n 1)3 n 4. The following pattern emerges for the probability space for repeatedly rolling an n-sided die until a 1 is rolled: (n 1) i 1 P[roll a 1] i1 Since you roll first and turns alternate, we are interested in the cases above in which index i {0, 2, 4,...}, so we have n i P[win] i0 (n 1) 2i n 2i+1 9. [6 POINTS] How many students have to be in the classroom to guarantee that two of them are born in the same month? ANSWER: This is a simple application of the pigeonhole principle. Once we have 13 students in the room, we are guaranteed to have two of them born in the same month. 10. [6 POINTS] A given probability space has five outcomes Ω {ω 1, ω 2, ω 3, ω 4, ω 5 } with probabilities P 1 2p, P 2 3p, P 3 4p, P 4 4p, P 5 2p. What is p? ANSWER: The probilities must sum to 1; therefore, we have 1 P 1 + P 2 + P 3 + P 4 + P 5 2p + 3p + 4p + 4p + 2p 15p Solving the above for p, our answer is p

4 11. [6 POINTS] Half of all students use emacs. The other half use vi. One in 10 vi users gets an A, while only one in 100 emacs users gets an A. You select a student at random; this student got an A. What is the probability that this student uses vi? ANSWER: Here, we are calculating P[student uses vi student got an A]. Using the conditional probability equation with A representing student uses vi and B representing student got an A, we have P[A B] P[A B] P[B] First, for a population of say 200, we have P[A B] Then, we have P[A B] P[A B] P[B] and P[B] [6 POINTS] A valid license plate consists of three uppercase letters (A-Z) followed by four digits (0-9). The digits must be in non-decreasing order and must contain exactly one 7. Examples of valid license plates include YAY1788, YYZ0079, VIM1117, etc. How many valid license plates are possible? ANSWER: First, for the three uppercase letters, these are selected independently. Therefore, there are possibilities. Next, for the four digits, we fix exactly one 7. Then, this problem is asking you to count the number of three-digit sequences in which each subsequent digit going from left to right does not decrease. To simplify the representation, we use delimiters between each possible digit not including 7. Let each represent a digit, and each represent a change from one digit to the next digit in the set {0, 1, 2, 3, 4, 5, 6, 8, 9}. There will always be exactly three characters (because we require a three-digit sequence) and exactly eight delimiters, for an overall total of 11 characters. Given this representation, we need to count the number of ways we can place eight delimiters before/among/after three digits. To accomplish this, we need to choose three spots out of 11. ( ) 11 Therefore, our answer is (Note that this totals , 900, 040.) 4

5 13. [6 POINTS] How many numbers in the set {1, 2, 3,..., 500} are divisible by 3 or 5? ANSWER: This is an inclusion-exclusion problem. First, calculate how many numbers are divisible by 3 as Next, calculate how many numbers are divisible by 5 as Since there are some numbers that overlap here, we also calculate how many numbers are divisible by as Using inclusion-exclusion, our answer is [6 POINTS] How many non-isomorphic graphs can you construct with exactly four vertices, i.e., the set V {v 1, v 2, v 3, v 4 }? ANSWER: Two graphs G 1 and G 2 are isomorphic if we can relabel the vertices of G 2 to obtain G 1. In other words, the structure is the same, in particular the relationships/edges between vertices. With four vertices, there are 11 non-isomorphic graphs with degree sequences: δ [0, 0, 0, 0] ( 4 i1 δ i 0) δ [1, 1, 0, 0] ( 4 i1 δ i 2) δ [2, 1, 1, 0] ( 4 i1 δ i 4) δ [1, 1, 1, 1] ( 4 i1 δ i 4) δ [2, 2, 1, 1] ( 4 i1 δ i 6) δ [2, 2, 2, 0] ( 4 i1 δ i 6) δ [3, 1, 1, 1] ( 4 i1 δ i 6) δ [2, 2, 2, 2] ( 4 i1 δ i 8) δ [3, 2, 2, 1] ( 4 i1 δ i 8) δ [3, 3, 2, 2] ( 4 i1 δ i 10) δ [3, 3, 3, 3] ( 4 i1 δ i 12) Draw each of these and convince yourself that there are no other non-isomorphic graphs with exactly four vertices. 5

6 15. [8 POINTS] How many of the 5-digit sequences through have digits that sum exactly to 15? ANSWER: The key to solving this problem is representation (and inclusion-exclusion). To simplify the representation, we use a character to represent one unit. We need 15 such units to obtain a sum of 15. Since we require a five-digit sequence, we use four delimiters. As an example, represents (and ). There will always be exactly 15 characters and exactly four delimiters, for an overall total of 19 characters. Given this representation, we need to count the number of ways we can place four delimiters before/among/after 19 characters. To accomplish this, we need to choose four spots out of 19. Unfortunately, not all solutions using this representation are valid. If we aggregate more than nine characters, we have an invalid solution (e.g., represents?0212, which is invalid. We must therefore exclude all possible ways in which we can have 10 adjacent characters. We can calculate this by counting the ways in which we can place the four delimiters before/among/after characters. Therefore, our answer is ( 19 4 ) 5 ( )

What is the probability that Professor Goldschmidt gets locked up (at position 3) rather than reaching home (at position 0)? ANSWER: This is the random walk problem.

7 16. [8 POINTS] After losing the vi versus emacs showdown, Professor Goldschmidt gets drunk. He leaves the bar at position 1 (shown in the diagram below) and takes independent steps: left (L) with probability 2 3 or right (R) with probability 1 3. What is the probability that Professor Goldschmidt gets locked up (at position 3) rather than reaching home (at position 0)? ANSWER: This is the random walk problem. There are (at least) two possible approaches to solving this problem: (i) Use the law of total probability to calculate P[jail] as follows: P[jail] P[jail L] P [L] + P[jail RL] P [RL] + P[jail RR] P [RR] ( 2 ( 1 (0) + P[jail] 3) 3 3) 2 ( 1 + (1) 3 1 3) P [jail] Solving the above, we find P[jail] 1 7 (thank goodness). (ii) We can identify all possible sequences that lead Professor Goldschmidt to the lockup as RR, RLRR, RLRLRR,..., (RL) i RR,... (i 0) Since each step is independent, we first observe that each possible sequence consists of P[RR] and each occurrence of P[RL] Again since each step is independent, adding up the probabilities for each sequence that leads to the lockup yields P[jail] P[RR] + P[RLRR] + P[RLRLRR] +... ) ( 1 ( 2 2 ) + 9 9) ( i0 1 2 ) i 1 9( 9 9 i0 ( 2 9) i

Counting Methods. CSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo

Counting Methods. CSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo Counting Methods CSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo c Xin He (University at Buffalo) CSE 191 Discrete Structures 1 / 48 Need for Counting The problem of counting