STAT 418HW1 SolutionsSections 1.1-1.53,5,7,11,14,21,22,27; TE 8,10,123. Consider each job as a place in a sequence. Because only one person canbe assigned to a job, it follows that the total number of possible assignmentsis a permutation of the numbers 1, . . . , 20. There are 20! possible differentassignments.5. Because the order matters in an area code, by the multiplication rule wemay simply multiply the number of possibilities in each place: 8*2*9 = 144.If the first digit must be a 4, there are 1*2*9 = 18 possible area codes.7.a. Because each child is distinct, the order matters and there are 6! =720 permutations of the 3 boys and girls.b. There are 2! ways to choose which gender is first in the row. There are3! ways to order the boys and 3! ways to order the girls. Therefore,there are 2!3!3! = 72 ways 3 boys and 3 girls can sit in a row if bothgenders sit together.c. If only the boys sit together, there are 4 ways to choose where they sitamongst the girls. There are 3! ways to order the boys and 3! waysto order the girls. Therefore, there are 4*3!3! = 144 ways to seat thechildren if only the boys must sit together.d. There are 3! ways to order the boys and 3! ways to order the girls.If no two people of the same sex may sit together, then they mustalternate and there are 2 ways to choose which gender is in the firstseat. There are 2*3!3! = 72 ways to seat the children if no 2 peopleof the same sex are allowed to sit together.11.a. There are 6 books. Because the order matters, there 6! = 720 possiblepermutations.b. There are 3 types of books and 3! ways to order the type of book.There are 3! ways to order the novels and 2! ways to order the mathbooks. There are 3!3!2! = 72 ways to order the books.1c. There are 3! ways to order the novels and 3! ways to order the mathand chemistry books. There are 4 ways to choose where the novelswill be placed among the other books, so there 4*3!3! = 144 ways toorder the books.14. Because the order of your cards does not matter in a poker hand, thereare525=52·51·50·49·485·4·3·2= 2, 598, 960 5-card poker hands.21. Consider the 7 possible moves, RRRRUUU. Each path is identified bythe order of the R’s and U’s, where the R’s are alike and the U’s are alike.There are7!4!3!=7·6·53!= 35 different paths from A to B.22. There are4!2!2!ways to get to the circled point from A. There are3!2!1!ways to get to B from the circled point. By the Multiplication Rule, thereare4!2!2!·3!2!1!= 6 · 3 = 18 ways to get from A to B by going through thecircled point.27. Because the 12 people are distinct, there are123, 4, 5=12!3!4!5!=11·10·9·8·72= 27, 720 possible divisions of 12 people into committees of sizes3, 4, and 5.TE 8. Consider a group of n men and a group of m women. There arenrm0ways to choose r men. There arenr − 1m1ways to chooser − 1 men and 1 woman. Similarly, there arenkmr − kways to choose kmen and r − k women, k = 0, . . . , r. Because each of these ways is distinct,there arePrk=0nkmr − kways to choose a group of r objects.TE 10.a. There arenkways to choose the committee of k from the group of npeople. There are k choices for the chair. By the Multiplication Rule,there arenkk ways to choose the committee and chair.b. There arenk − 1ways to choose the committee of k − 1 from thegroup of n people. There are n − (k − 1) = n − k + 1 choices for the2chair. By the Multiplication Rule, there arenk − 1(n − k + 1) waysto choose the committee and chair.c. There are n ways to choose the chair. There aren − 1k − 1ways tochoose the committee. By the Multiplication Rule, there aren − 1k − 1nways to choose the committee and chair.d. In all ways, we choose a committee and a chair; therefore each calcu-lation must be equal.e.knk=k · n!(n − k)!k!=n!(n − k)!(k − 1)!=(n − k + 1) · n!(n − k + 1)!(k − 1)!= (n − k + 1)nk − 1.knk=k · n · (n − 1)!(n − 1 − k + 1)!k · (k − 1)!= n(n − 1)!((n − 1) − (k − 1))!(k − 1)!= nn − 1k − 1.TE 12.a. (i) For k = 1, . . . , n, there are knkways to choose a committeeand a chair. Because each size is distinct, there arePnk=1knkways to choose a committee of any size and a chair.(ii) There are n possible ways to choose a chair. For every otherperson, they are either chosen or not, so there are 2n−1waysto choose the remaining members. By the Multiplication Rule,3there are n2n−1ways to choose a committee of any size and achair.b. (i) There arenkways to choose a committee of k from n people.There are k ways to choose the chair. Because the secretary andchair can be the same person, there are also k ways to choosethe secretary. For k = 1, . . . , n, there are k2nkways to choosethe committee, chair, and secretary. Because each size is distinct,there arePnk=1k2nkways to choose the committee of any size,chair, and secretary.(ii) There are n2n−1ways to choose the committee of any size andhave the same person be the chair and secretary. There aren(n − 1)2n−2ways to choose so they are different. n2n−1+ n(n −1)2n−2= n2n−2(2 + n − 1) = n2n−2(n + 1).c. (i) There arenkways to choose a committee of k from n people.There are k ways to choose the chair. Because the secretary andchair can be the same person, there are also k ways to choose thesecretary. If we also allow the treasurer to be the same person asthe secretary and/or chair, there are k ways to choose a treasurer.For k = 1, . . . , n, there are k3nkways to choose the committee,chair, secretary, and treasurer. Because each size is distinct, therearePnk=1k3nkways to choose the committee of any size, chair,secretary, and treasurer.(ii) There are n2n−1ways to choose the committee of any size andhave the same person be the chair, secretary, and treasurer. Thereare 3n(n − 1)2n−2ways to choose so two are the same and one isdifferent. There are n(n − 1)(n − 2)2n−3ways to choose so theyare all different. n2n−1+ 3n(n − 1)2n−2+ n(n − 1)(n − 2)2n−3=n2n−3(2 ∗ 2 + 3 ∗ 2(n − 1) + (n − 1)(n − 2)) = n2n−3(4 + 6n − 6 +n2− 3n + 2) = n2n−3(n2+ 3n) = 2n−3n2(n +