Exam (elaborations) TEST BANK FOR A First Course in Probability 8th Edition By Sheldon Ross (Solution Manual)

Exam (elaborations) TEST BANK FOR A First Course in Probability 8th Edition By Sheldon Ross (Solution Manual) A Solution Manual for: A First Course In Probability by Sheldon M. Ross. John L. Weatherwax February 7, 2012 Introduction Here you’ll find some notes that I wrote up as I worked through this excellent book. I’ve worked hard to make these notes as good as I can, but I have no illusions that they are perfect. If you feel that that there is a better way to accomplish or explain an exercise or derivation presented in these notes; or that one or more of the explanations is unclear, incomplete, or misleading, please tell me. If you find an error of any kind – technical, grammatical, typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments in later printings the name of the first person to bring each problem to my attention. Acknowledgements Special thanks to (most recent comments are listed first): Mark Chamness, Dale Peterson, Doug Edmunds, Marlene Miller, JohnWilliams (several contributions to chapter 4), Timothy Alsobrooks, Konstantinos Stouras, William Howell, Robert Futyma, Waldo Arriagada, Atul Narang, Andrew Jones, Vincent Frost, and Gerardo Robert for helping improve these notes and solutions. It should be noted that Marlene Miller made several helpful suggestions on most of the material in Chapter 3. Her algebraic use of event “set” notation to solve probability problems has opened my eyes to this powerful technique. It is a tool that I wish to become more proficient with. All comments (no matter how small) are much appreciated. In fact, if you find these notes useful I would appreciate a contribution in the form of a solution to a problem that is not yet  1 worked in these notes. Sort of a “take a penny, leave a penny” type of approach. Remember: pay it forward. Miscellaneous Problems The Crazy Passenger Problem The following is known as the “crazy passenger problem” and is stated as follows. A line of 100 airline passengers is waiting to board the plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let’s say that the k-th passenger in line has a ticket for the seat number k.) Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)? If one tries to solve this problem with conditional probability it becomes very difficult. We begin by considering the following cases if the first passenger sits in seat number 1, then all the remaining passengers will be in their correct seats and certainly the #100’th will also. If he sits in the last seat #100, then certainly the last passenger cannot sit there (in fact he will end up in seat #1). If he sits in any of the 98 seats between seats #1 and #100, say seat k, then all the passengers with seat numbers 2, 3, . . . , k−1 will have empty seats and be able to sit in their respective seats. When the passenger with seat number k enters he will have as possible seating choices seat #1, one of the seats k + 1, k +2, . . . , 99, or seat #100. Thus the options available to this passenger are the same options available to the first passenger. That is if he sits in seat #1 the remaining passengers with seat labels k+1, k+2, . . . , 100 can sit in their assigned seats and passenger #100 can sit in his seat, or he can sit in seat #100 in which case the passenger #100 is blocked, or finally he can sit in one of the seats between seat k and seat #99. The only difference is that this k-th passenger has fewer choices for the “middle” seats. This k passenger effectively becomes a new “crazy” passenger. From this argument we begin to see a recursive structure. To fully specify this recursive structure lets generalize this problem a bit an assume that there are N total seats (rather than just 100). Thus at each stage of placing a k-th crazy passenger we can choose from • seat #1 and the last or N-th passenger will then be able to sit in their assigned seat, since all intermediate passenger’s seats are unoccupied. • seat # N and the last or N-th passenger will be unable to sit in their assigned seat. • any seat before the N-th and after the k-th. Where the k-th passenger’s seat is taken by a crazy passenger from the previous step. In this case there are N −1−(k+1)+1 = N − k − 1 “middle” seat choices. If we let p(n, 1) be the probability that given one crazy passenger and n total seats to select from that the last passenger sits in his seat. From the argument above we have a recursive structure give by p(N, 1) = 1 N (1) + 1 N (0) + 1 N NX−1 k=2 p(N − k, 1) = 1 N + 1 N NX−1 k=2 p(N − k, 1) . where the first term is where the first passenger picks the first seat (where the N will sit correctly with probability one), the second term is when the first passenger sits in the N-th seat (where the N will sit correctly with probability zero), and the remaining terms represent the first passenger sitting at position k, which will then require repeating this problem with the k-th passenger choosing among N − k + 1 seats. To solve this recursion relation we consider some special cases and then apply the principle of mathematical induction to prove it. Lets take N = 2. Then there are only two possible arrangements of passengers (1, 2) and (2, 1) of which one (the first) corresponds to the second passenger sitting in his assigned seat. This gives p(2, 1) = 1 2 . If N = 3, then from the 3! = 6 possible choices for seating arrangements (1, 2, 3) (1, 3, 2) (2, 3, 1) (2, 1, 3) (3, 1, 2) (3, 2, 1) Only (1, 2, 3) (2, 1, 3) (3, 2, 1) correspond to admissible seating arrangements for this problem so we see that p(3, 1) = 3 6 = 1 2 . If we hypothesis that p(N, 1) = 1 2 for all N, placing this assumption into the recursive formulation above gives p(N, 1) = 1 N + 1 N NX−1 k=2 1 2 = 1 2 . Verifying that indeed this constant value satisfies our recursion relationship. Chapter 1 (Combinatorial Analysis) Chapter 1: Problems Problem 1 (counting license plates) Part (a): In each of the first two places we can put any of the 26 letters giving 262 possible letter combinations for the first two characters. Since the five other characters in the license plate must be numbers, we have 105 possible five digit letters their specification giving a total of 262 · 105 = , total license plates. Part (b): If we can’t repeat a letter or a number in the specification of a license plate then the number of license plates becomes 26 · 25 · 10 · 9 · 8 · 7 · 6 = , total license plates. Problem 2 (counting die rolls) We have six possible outcomes for each of the die rolls giving 64 = 1296 possible total outcomes for all four rolls. Problem 3 (assigning workers to jobs) Since each job is different and each worker is unique we have 20! different pairings. Problem 4 (creating a band) If each boy can play each instrument we can have 4! = 24 ordering. If Jay and Jack can play only two instruments then we will assign the instruments they play first with 2! possible orderings. The other two boys can be assigned the remaining instruments in 2! ways and thus we have 2! · 2! = 4 , possible unique band assignments. Problem 5 (counting telephone area codes) In the first specification of this problem we can have 9 − 2 + 1 = 8 possible choices for the first digit in an area code. For the second digit there are two possible choices. For the third digit there are 9 possible choices. So in total we have 8 · 2 · 9 = 144 , possible area codes. In the second specification of this problem, if we must start our area codes with the digit “four” we will only have 2 · 9 = 18 area codes. Problem 6 (counting kittens) The traveler would meet 74 = 2401 kittens. Problem 7 (arranging boys and girls) Part (a): Since we assume that each person is unique, the total number of ordering is given by 6! = 720. Part (b): We have 3! orderings of each group of the three boys and girls. Since we can put these groups of boys and girls in 2! different ways (either the boys first or the girls first) we have (2!) · (3!) · (3!) = 2 · 6 · 6 = 72 , possible orderings. Part (c): If the boys must sit together we have 3! = 6 ways to arrange the block of boys. This block of boys can be placed either at the ends or in between any of the individual 3! orderings of the girls. This gives four locations where our block of boys can be placed we have 4 · (3!) · (3!) = 144 , possible orderings. Part (d): The only way that no two people of the same sex can sit together is to have the two groups interleaved. Now there are 3! ways to arrange each group of girls and boys, and to interleave we have two different choices for interleaving. For example with three boys and girls we could have g1b1g2b2g3b3 vs. b1g1b2g2b3g3 , thus we have 2 · 3! · 3! = 2 · 62 = 72 , possible arrangements. Problem 8 (counting arrangements of letters) Part (a): Since “Fluke” has five unique letters we have 5! = 120 possible arrangements. Part (b): Since “Propose” has seven letters of which four (the “o”’s and the “p”’s) repeat we have 7! 2! · 2! = 1260 , arrangements. Part (c): Now “Mississippi” has eleven characters with the “i” repeated four times, the “s” repeated four times and the “p” repeated two times, so we have 11! 4! · 4! · 2! = 34650 , possible rearranges. Part (d): “Arrange” has seven characters with a double “a” and a double “r” so it has 7! 2! · 2! = 1260 , different arrangements. Problem 9 (counting colored blocks) Assuming each block is unique we have 12! arrangements, but since the six black and the four red blocks are not distinguishable we have 12! 6! · 4! = 27720 , possible arrangements. Problem 10 (seating people in a row) Part (a): We have 8! = 40320 possible seating arrangements. Part (b): We have 6! ways to place the people (not including A and B). We have 2! ways to order A and B. Once the pair of A and B is determined, they can be placed in between any ordering of the other six. For example, any of the “x”’s in the expression below could be replaced with the A B pair x P1 x P2 x P3 x P4 x P5x P6 x . Giving seven possible locations for the A,B pair. Thus the total number of orderings is given by 2! · 6! · 7 = 10080 . Part (c): To place the men and women according to the given rules, the men and women must be interleaved. We have 4! ways to arrange the men and 4! ways to arrange the women. We can start our sequence of eight people with a woman or a man (giving two possible choices). We thus have 2 · 4! · 4! = 1152 , possible arrangements. Part (d): Since the five men must sit next to each other their ordering can be specified in 5! = 120 ways. This block of men can be placed in between any of the three women, or at the end of the block of women, who can be ordered in 3! ways. Since there are four positions we can place the block of men we have 5! · 4 · 3! = 2880 , possible arrangements. Part (e): The four couple have 2! orderings within each pair, and then 4! orderings of the pairs giving a total of (2!)4 · 4! = 384 , total orderings. Problem 11 (counting arrangements of books) Part (a): We have (3 + 2 + 1)! = 6! = 720 arrangements. Part (b): The mathematics books can be arranged in 2! ways and the novels in 3! ways. Then the block ordering of mathematics, novels, and chemistry books can be arranged in 3! ways resulting in (3!) · (2!) · (3!) = 72 , possible arrangements. Part (c): The number of ways to arrange the novels is given by 3! = 6 and the other three books can be arranged in 3! ways with the blocks of novels in any of the four positions in between giving 4 · (3!) · (3!) = 144 , possible arrangements. Problem 12 (counting awards) Part (a): We have 30 students to choose from for the first award, and 30 students to choose from for the second award, etc. So the total number of different outcomes is given by 305 = Part (b): We have 30 students to choose from for the first award, 29 students to choose from for the second award, etc. So the total number of different outcomes is given by 30 · 29 · 28 · 27 · 26 = Problem 13 (counting handshakes) With 20 people the number of pairs is given by  20 2  = 190 . Problem 14 (counting poker hands) A deck of cards has four suits with thirteen cards each giving in total 52 cards. From these 52 cards we need to select five to form a poker hand thus we have  52 5  = , unique poker hands. Problem 15 (pairings in dancing) We must first choose five women from ten in  10 5  possible ways, and five men from 12 in  12 5  ways. Once these groups are chosen then we have 5! pairings of the men and women. Thus in total we will have  10 5   12 5  5! = 252 · 792 · 120 = , possible pairings. Problem 16 (forced selling of books) Part (a): We have to select a subject from three choices.  If we choose math we have 6 2  = 15 choices of books to sell. If we choose science we have  7 2  = 21 choices of books to sell. If we choose economics we have  4 2  = 6 choices of books to sell. Since each choice is mutually exclusive in total we have 15 + 21 + 6 = 42, possible choices. Part (b): We must pick two subjects from  3 2  = 3 choices. If we denote the letter “M” for the choice math the letter “S” for the choice science, and the letter “E” for the choice economics then the three choices are (M, S) (M,E) (S,E) . For each of the choices above we have 6 · 7 + 6 · 4 + 7 · 4 = 94 total choices. Problem 17 (distributing gifts) We can choose seven children to give gifts to in  10 7  ways. Once we have chosen the seven children, the gifts can be distributed in 7! ways. This gives a total of  10 7  · 7! = , possible gift distributions. Problem 18 (selecting political parties) We can choose two Republicans from the five total in  5 2  ways, we can choose two Democrats from the six in  6 2  ways, and finally we can choose three Independents from the four in  4 3  ways. In total, we will have  5 2  ·  6 2  ·  4 3  = 600 , different committees. Problem 19 (counting committee’s with constraints) Part (a): We select three men from six in  6 3  , but since two men won’t serve together we need to compute the number of these pairings of three men that have the two that won’t serve together. The number of committees we can form (with these two together) is given by  2 2  ·  4 1  = 4 . So we have  6 3  − 4 = 16 , possible groups of three men. Since we can choose  8 3  = 56 different groups of women, we have in total 16 · 56 = 896 possible committees. Part (b): If two women refuse to serve together, then we will have  2 2  ·  6 1  groups with these two women in them from the  8 3  ways to draw three women from eight. Thus we have  8 3  −  2 2  ·  6 1  = 56 − 6 = 50 , possible groupings of woman. We can select three men from six in  6 3  = 20 ways. In total then we have 50 · 20 = 1000 committees. Part (c): We have  8 3  ·  6 3  total committees, and  1 1  ·  7 2  ·  1 1  ·  5 2  = 210 , committees containing the man and women who refuse to serve together. So we have  8 3  ·  6 3  −  1 1  ·  7 2  ·  1 1  ·  5 2  = 1120 − 210 = 910 , total committees. Problem 20 (counting the number of possible parties) Part (a): There are a total of  8 5  possible groups of friends that could attend (assuming no feuds). We have  2 2  ·  6 3  sets with our two feuding friends in them, giving  8 5  −  2 2  ·  6 3  = 36 possible groups of friends Part (b): If two fiends must attend together we have that  2 2  6 3  if the do attend the party together and  6 5  if they don’t attend at all, giving a total of  2 2  6 3  +  6 5  = 26 . Problem 21 (number of paths on a grid) From the hint given that we must take four steps to the right and three steps up, we can think of any possible path as an arraignment of the letters ”U” for up and “R” for right. For example the string U U U RRRR, would first step up three times and then right four times. Thus our problem becomes one of counting the number of unique arrangements of three “U”’s and four “R”’s, which is given by 7! 4! · 3! = 35 . Problem 22 (paths on a grid through a specific point) One can think of the problem of going through a specific point (say P) as counting the number of paths from the start A to P and then counting the number of paths from P to the end B. To go from A to P (where P occupies the (2, 2) position in our grid) we are looking for the number of possible unique arrangements of two “U”’s and two “R”’s, which is given by 4! 2! · 2! = 6 , possible paths. The number of paths from the point P to the point B is equivalent to the number of different arrangements of two “R”’s and one “U” which is given by 3! 2! · 1! = 3 . From the basic principle of counting then we have 6 · 3 = 18 total paths. Problem 23 (assignments to beds) Assuming that twins sleeping in different bed in the same room counts as a different arraignment, we have (2!) · (2!) · (2!) = 8 possible assignments of each set of twins to a room. Since there are 3! ways to assign the pair of twins to individual rooms we have 6 · 8 = 48 possible assignments. Problem 24 (practice with the binomial expansion) This is given by (3x2 + y)5 = X5 k=0  5 k  (3x2)ky5−k . Problem 25 (bridge hands) We have 52! unique permutations, but since the different arrangements of cards within a given hand do not matter we have 52! (13!)4 , possible bridge hands. Problem 26 (practice with the multinomial expansion) This is given by the multinomial expansion (x1 + 2x2 + 3x3)4 = X n1+n2+n3=4  4 n1 , n2 , n3  xn1 1 (2x2)n2(3x3)n3 The number of terms in the above summation is given by  4 + 3 − 1 3 − 1  =  6 2  = 6 · 5 2 = 15 . Problem 27 (counting committees) This is given by the multinomial coefficient  12 3 , 4 , 5  = 27720 Problem 28 (divisions of teachers) If we decide to send n1 teachers to school one and n2 teachers to school two, etc. then the total number of unique assignments of (n1, n2, n3, n4) number of teachers to the four schools is given by  8 n1 , n2 , n3 , n4  . Since we want the total number of divisions, we must sum this result for all possible combinations of ni, or X n1+n2+n3+n4=8  8 n1 , n2 , n3 , n4  = (1 + 1 + 1 + 1)8 = 65536 , possible divisions. If each school must receive two in each school, then we are looking for  8 2 , 2 , 2 , 2  = 8! (2!)4 = 2520 , orderings. Problem 29 (dividing weight lifters) We have 10! possible permutations of all weight lifters but the permutations of individual countries (contained within this number) are irrelevant. Thus we can have 10! 3! · 4! · 2! · 1! =  10 3 , 4 , 2 , 1  = 12600 , possible divisions. If the united states has one competitor in the top three and two in the bottom three. We have  3 1  possible positions for the US member in the first three positions and  3 2  possible positions for the two US members in the bottom three positions, giving a total of  3 1  3 2  = 3 · 3 = 9 , combinations of US members in the positions specified. We also have to place the other countries participants in the remaining 10 − 3 = 7 positions. This can be done in  7 4 , 2 , 1  = 7! 4!·2!·1! = 105 ways. So in total then we have 9 · 105 = 945 ways to position the participants. Problem 30 (seating delegates in a row) If the French and English delegates are to be seated next to each other, they can be can be placed in 2! ways. Then this pair constitutes a new “object” which we can place anywhere among the remaining eight people, i.e. there are 9! arrangements of the eight remaining people and the French and English pair. Thus we have 2·9! = possible combinations. Since in some of these the Russian and US delegates are next to each other, this number over counts the true number we are looking for by 2 · 28! = (the first two is for the number of arrangements of the French and English pair). Combining these two criterion we have 2 · (9!) − 4 · (8!) = . Problem 31 (distributing blackboards) Let xi be the number of black boards given to school i, where i = 1, 2, 3, 4. Then we must have P i xi = 8, with xi  0. The number of solutions to an equation like this is given by  8 + 4 − 1 4 − 1  =  11 3  = 165 . If each school must have at least one blackboard then the constraints change to xi  1 and the number of such equations is give by  8 − 1 4 − 1  =  7 3  = 35 . Problem 32 (distributing people) Assuming that the elevator operator can only tell the number of people getting off at each floor, we let xi equal the number of people getting off at floor i, where i = 1, 2, 3, 4, 5, 6. Then the constraint that all people are off at the sixth floor means that P i xi = 8, with xi  0. This has  n + r − 1 r − 1  =  8 + 6 − 1 6 − 1  =  13 5  = 1287 , possible distribution people. If we have five men and three women, let mi and wi be the number of men and women that get off at floor i. We can solve this problem as the combination of two problems. That of tracking the men that get off on floor i and that of tracking the women that get off on floor i. Thus we must have X6 i=1 mi = 5 mi  0 X6 i=1 wi = 3 wi  0 . The number of solutions to the first equation is given by  5 + 6 − 1 6 − 1  =  10 5  = 252 , while the number of solutions to the second equation is given by  3 + 6 − 1 6 − 1  =  8 5  = 56 . So in total then (since each number is exclusive) we have 252 · 56 = 14114 possible elevator situations. Problem 33 (possible investment strategies) Let xi be the number of investments made in opportunity i. Then we must have X4 i=1 xi = 20 with constraints that x1  2, x2  2, x3  3, x4  4. Writing this equation as x1 + x2 + x3 + x4 = 20 we can subtract the lower bound of each variable to get (x1 − 2) + (x2 − 2) + (x3 − 3) + (x4 − 4) = 20 − 2 − 2 − 3 − 4 = 9 . Then defining v1 = x1 − 2, v2 = x2 − 2, v3 = x3 − 3, and v4 = x4 − 4, then our equation becomes v1 + v2 + v3 + v4 = 9, with the constraint that vi  0. The number of solutions to equations such as these is given by  9 + 4 − 1 4 − 1  =  12 3  = 220 . Part (b): First we pick the three investments from the four possible in  4 3  = 4 possible ways. The four choices are denoted in table 1, where a one denotes that we invest in that option. Then investment choice number one requires the equation v2+v3+v4 = 20−2−3−4 = choice v1 = x1 − 2  0 v2 = x2 − 2  0 v3 = x3 − 3  0 v4 = x4 − 4  0 1 0 1 1 1 2 1 0 1 1 3 1 1 0 1 4 1 1 1 0 Table 1: All possible choices of three investments. 11, and has  11 + 3 − 1 3 − 1  =  13 2  = 78 possible solutions. Investment choice number two requires the equation v1+v3+v4 = 20−2−3−4 = 11, and again has  11 + 3 − 1 3 − 1  =  13 2  = 78 possible solutions. Investment choice number three requires the equation v1+v2+v4 = 20−2−2−4 = 12, and has  12 + 3 − 1 3 − 1  =  14 2  = 91 possible solutions. Finally, investment choice number four requires the equation v1+v2+v3 = 20−2−2−3 = 13, and has  13 + 3 − 1 3 − 1  =  15 2  = 105 possible solutions. Of course we could also invest in all four opportunities which has the same number of possibilities as in part (a) or 220. Then in total since we can do any of these choices we have 220 + 105 + 91 + 78 + 78 = 572 choices. Chapter 1: Theoretical Exercises Problem 1 (the generalized counting principle) This can be proved by recursively applying the basic principle of counting. Problem 2 (counting dependent experimental outcomes) We have m choices for the outcome of the first experiment. If the first experiment returns i as an outcome, then there are ni possible outcomes for the second experiment. Thus if the experiment returns “one” we have n1 possible outcomes, if it returns “two” we have n2 possible outcomes, etc. To count the number of possible experimental outcomes we can envision a tree like structure representing the totality of possible outcomes, where we have m branches leaving the root node indicating the m possible outcomes from the first experiment. From the first of these branches we have n1 additional branches representing the outcome of the second experiment when the first experimental outcome was a one. From the second branch we have n2 additional branches representing the outcome of the second experiment when the first experimental outcome was a two. We can continue this process, with the m-th branch from the root node having nm leaves representing the outcome of the second experiment when the first experimental outcome was a m. Counting all of these outcomes we have n1 + n2 + n3 + · · · + nm , total experimental outcomes. Problem 3 (selecting r objects from n) To select r objects from n, we will have n choices for the first object, n − 1 choices for the second object, n − 2 choices for the third object, etc. Continuing we will have n − r + 1 choices for the selection of the r-th object. Giving a total of n(n − 1)(n − 2) · · ·(n − r + 1) total choices if the order of selection matters. If it does not then we must divide by the number of ways to rearrange the r selected objects i.e. r! giving n(n − 1)(n − 2) · · ·(n − r + 1) r! , possible ways to select r objects from n when the order of selection of the r object does not matter. Problem 4 (combinatorial explanation of  n k  ) If all balls are distinguishable then there are n! ways to arrange all the balls. With in this arrangement there are r! ways to uniquely arrange the black balls and (n − r)! ways to uniquely arranging the white balls. These arrangements don’t represent new patterns since the balls with the same color are in fact indistinguishable. Dividing by these repeated patterns gives n! r!(n − r)! , gives the unique number of permutations. Problem 5 (the number of binary vectors who’s sum is greater than k ) To have the sum evaluate to exactly k, we must select at k components from the vector x to have the value one. Since there are n components in the vector  x, this can be done in n k  ways. To have the sum exactly equal k + 1 we must select k + 1 components from x to have a value one. This can be done in  n k + 1  ways. Continuing this pattern we see that the number of binary vectors x that satisfy Xn i=1 xi  k is given by Xn l=k  n l  =  n n  +  n n − 1  +  n n − 2  + . . . +  n k + 1  +  n k  . Problem 6 (counting the number of increasing vectors) If the first component x1 were to equal n, then there is no possible vector that satisfies the inequality x1 < x2 < x3 < . . . < xk constraint. If the first component x1 equals n − 1 then again there are no vectors that satisfy the constraint. The first largest value that the component x1 can take on and still result in a complete vector satisfying the inequality constraints is when x1 = n−k+1 For that value of x1, the other components are determined and are given by x2 = n − k + 2, x3 = n − k + 3, up to the value for xk where xk = n. This assignment provides one vector that satisfies the constraints. If x1 = n − k, then we can construct an inequality satisfying vector x by assigning the k − 1 other components x2, x3, up to xk by assigning the integers n − k + 1 , n − k + 2 , . . . n − 1 , n to the k − 1 components. This can be done in  k 1  ways. Continuing if x1 = n − k − 1, then we can obtain a valid vector x by assign the integers n − k , n − k + 1 , . . . n − 1 , n to the k − 1 other components of x. This can be seen as an equivalent problem to that of specifying two blanks from n − (n − k) + 1 = k + 1 spots and can be done in  k + 1 2  ways. Continuing to decrease the value of the x1 component, we finally come to the case where we have n locations open for assignment with k assignments to be made (or equivalently n − k blanks to be assigned) since this can be done in  n n − k  ways. Thus the total number of vectors is given by 1 +  k 1  +  k + 1 2  +  k + 2 3  + . . . +  n − 1 n − k − 1  +  n n − k  . Problem 7 (choosing r from n by drawing subsets of size r − 1) Equation 4.1 from the book is given by  n r  =  n − 1 r − 1  +  n − 1 r