ECE-316 Tutorial for the week May 11-15 Problem 4: Page 16 - (Refer to lecture notes part 1- slide 14,16) John, Jim, Jay and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What is John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums. If each of the boys can play all 4 instruments, there are 4! = 24 possible arrangements. By first assigning the piano and drum to Jay and Jack, there are 2 2 1 possibilities. John and Jim can only play the remaining 2 instruments with 2 2 1 possibilities. Therefore, there are total of (2. 2=4) possibilities. Problem 7: Page 16 - (Refer to lecture notes part 1- slide 16) (a) In how many ways can 3 boys and 3 girls sit in a row? (b) In how many ways can 3 boys and 3 girls sit in a row if the boys and the girls are each to sit together? (c) In how many ways if only boys must sit together? (d) In how many ways if no two people of the same sex are allowed to sit together?
a) Since there is no restriction of boys and girls sitting way, therefore we have a permutation of 6 people which is 6! = 720 possible ways. b) We have two separate groups (boys group and girls group) Number of combinations for boys group = 3! Number of combinations for girls group = 3! Now we have 2 items (boys group and girls group), In this case there are 2! possible combinations for placing these 2 groups. Total number of combinations = 3!. 3!. 2! = 72 c) Only boys must set together Number of combinations for boys group =3! Now we have group of 4 items (girl1, girl2, girl3, and boy s group), in this case there are (4!) possible combinations for placing these 4 items. Total number of combinations = 3!. 4! = 144 d) The fist position can be taken by anyone from boys or girls (6 people). The second position can be taken by one of the three people of the opposite sex from the first one. The third position can be taken by one of the two with the same sex as the first one. Then the fourth position has two possibilities from the opposite sex. And finally there are two people with opposite sex where can sit. Therefore, there are 6.3.2.2.1.1=72 possible ways of sitting. Another approach: No 2 people of same sex are allowed to sit together. Number of combinations for boys group = 3! Number of combinations for girls group = 3!
We can place these 2 groups in 2 possible ways B G B G B G or G B G B G B Total number of combinations = 3!. 3!. 2 = 72 Problem 15: Page 16 - (Refer to lecture notes part 1- slide 16) A dance class consists of 22 students, of which 10 are women and 12 are men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible? 10 5 possible ways to choose 5 women from group of 10 women. 12 5 possible ways to choose 5 men from group of 12 men. As for pairing, the first man can be paired with one of possible 5 women, the second man can be paired with one of possible 4 women, the third man can be paired with one of possible 3 women, the fourth man can be paired with one of two women and the fifth man is paired with the last woman, hence we have 5.4.3.2.1 = 5! combinations. Therefore, there are 10 12 5!.. 5 5 possible ways. Problem 16: Page 16 - (Refer to lecture notes part 1- slide 16) A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if (a) Both books are to be on the same subject? (b) Both books are to be on different subjects?
a) To choose 2 books from 6 math books, we have 6 2 possible ways. To choose 2 books from 7 science books, we have 7 2 possible ways. And to choose 2 books from 4 economic books, we have 4 2 possible ways. Therefore we have a total of 6 7 4 2 2 2 possibilities. b) We have (7. 6) possibilities when the books are chosen from science and math. We have (7. 4) possibilities when the books are chosen from science and economics. And we have (6. 4) possibilities when the books are chosen from math and economics. Therefore, we have a total of (7. 6 + 7. 4 + 6. 4) possibilities when the books are to be chosen from different books. Problem 17: Page 16 - (Refer to lecture notes part 1- slides 14,16) Seven different gifts are to be distributed among 10 children. How many distinct results are possible if no child is to receive more than one gift? We can do it in 2 different ways:
1) The first gift can go to any child of the 10 children, and the second can go to any child of the 9 children and the seventh gift can go to one of that last four children. So, there will be (10.9.8.7.6.5.4=604800) possibilities. OR 2) Assume that each child selects his gift Divide children into 2 groups, a group of 7 (each child gets one gift), and a group of 3 (each child gets NO gift), Number of possible combinations 10 7 Number of possible permutations of gifts within the group of 7 children = 7! Total number of possible combinations = 10.7! = 604800 7 Problem 19: Page 16 - (Refer to lecture notes part 1- slide 16) From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if (a) 2 of the men refuse to serve together? (b) 2 of the women refuse to serve together? (c) 1 man and 1 woman refuse to serve together?
a) 8 4 3 3 possible committees if the committee does not include the two men. And there are 8 4 2 3 21 possibilities if the committee includes one of the feuding men. Therefore there are total possibilities of 8 4 8 4 2 3 3 3 21 committees when 2 men refuse to serve together. b) 6 6 3 3 possible committees if the committees do not include the two women. And there are 6 26 213 possibilities if the committee includes one of the feuding women. Therefore there are total possibilities of 66 6 26 33 213 committees when 2 men refuse to serve together. c) 75 33 possible committees when neither feuding party serves. And when the feuding woman serves, there are 75 possible committees. 23
And when the feuding man serves, there are 7 5 3 2 possible committees. Therefore, there are total possibilities of 7 5 7 5 7 5 33 23 3 2 committees. Problem 29: Page 17 - (Refer to lecture notes part 1- slide 16) Ten weight lifters are competing in a team weight lifting contest. Of the lifters, 3 are from the United States, 4 are from Russia, 2 are from China, and 1 is from Canada. If the scoring takes account of the countries that the lifters represent, but not their individual identities, how many different outcomes are possible from the point of view of scores? How many different outcomes correspond to results in which the United States has 1 competitor in the top three and 2 competitors in the bottom three? 10 10! a) Number of possible different outcomes (ranking possibilities) = 3,4,2,1 3!.4!.2!.1! 12600 b) Given: USA has 1 competitor in the top 3 and 2 in the bottom 3. 3 possible combinations of having 1 USA in top 3. 1 3 possible combinations of having 2 USA in bottom 3. 2 7 7! possibilities of other 7 positions. 4,2,1 4!.2!.1! Total number of possible combinations = 3 3 7 1 2 4, 2,1 = 945