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CmSc 175 Discrete Mathematics Lesson 21: Solutions to Problems on Counting 1. Let A={a,b,c,d} a. Compute the number of permutations of three elements of A. P(4,3) = 4! / (4 – 3)! = 4! = 24 b. Compute the number of combinations of three elements of A. C(4,3) = 4! / (3! * (4 – 3)!) = 4! / 3! = 4 2. In how many ways can we select a chairperson, vice-chairperson and a recorder from a group of 11 persons? P(11,3) = 11! / 8! = 11.10.9 = 990 We have 11 choices for the chairperson. Once we choose the chairperson, we have 10 choices for the vice-chairperson (so far the number of choices is 11 * 10). Finally, we have 9 choices for the recorder, thus the number of choices is 11 * 10 * 9 = 990 3. In how many ways can we select a committee of three from a group of 11 persons? Here the order of the persons in the committee does not matter, so we have combinations of three elements out of 11: C(11,3) = 11! / (8! * 3!) = 11.10.9 / 3! = 990/6 = 165 In exercises 4-8 refer to a club consisting of 6 men and 7 women. 4. In how many ways can we select a committee of three men and four women? Here the order does not matter, so we have combinations of 3 men out of 6: C(6,3), and combinations of 4 women out of 7: C(7,4). For each selection of 3 men, we have all possible selections of 4 women, therefore the total number of selections is C(6,3) * C(7,4) C(6,3) = 6! / (3! * 3!) = (2*3*4*5*6) / 36 = 20 C(7,4) = 7! / (4! * 3!) = (2*3*4*5*6*7) / (6* 24) = 35 C(6,3) * C(7,4) = 20 * 35 = 700 1 5. In how many ways can we select a committee of four persons that has at least one woman? There are two ways to solve this problem. a) Consider all possible committees with at least one woman: Women Men Choices 1 3 C(7,1) * C(6,3) 2 2 C(7,2) * C(6,2) 3 1 C(7,3) * C(6,1) 4 0 C(7,4) * C(6,0) Then add all choices C(7,1) = 7 C(7,2) = 7! / (2! * 5!) = (2*3*4*5*6*7) / (2*2*3*4*5) = 21 C(7,3) = 7! / (3! * 4!) = (2*3*4*5*6*7) / (2*3*2*3*4) = 35 C(7,4) = C(7,3) = 35 C(6,3) = 6!/(3! * 3!) = (2*3*4*5*6) / (2*3*2*3) = 20 C(6,2) = 6! / (2! * 4!) = (2*3*4*5*6) / (2*2*3 * 4) = 15 C(6,1) = 6 C(6,0) = 1 C(7,1) * C(6,3) = 7 * 20 = 140 C(7,2) * C(6,2) = 21 * 15 = 315 C(7,3) * C(6,1) = 35 * 6 = 210 C(7,4) * C(6,0) = 35 * 1 = 35 ---------Total 700 b) All committees with at least one women are equal to the difference between all possible committees and committees with only men. All possible committees are C(13,4) = 13! (9! * 4!) = 10*11*12*13 / 2*3*4 = 5*11*13 = 715 Committees with only men are C(6,4) = 6! / (2! * 4!) = (2*3*4*5*6) / (2*2*3*4) = 15 Thus C(13,4) – C(6,4) = 715 – 15 = 700 2 6. In how many ways can we select a committee of four persons that has at most one man? Consider all possible committees with at most one man: Women Men Choices 3 1 C(7,3) * C(6,1) 4 0 C(7,4) * C(6,0) C(7,3) * C(6,1) = 210 (computed in problem 5) C(7,4) * C(6,0) = 35 Thus all possible committees with at most one man are 210 + 35 = 245 7. In how many ways can we select a committee of four persons that has persons of both sexes? We can solve the problem in two ways as illustrated in problem 5. Here we choose the second way – representing the choices as difference. The number of committees with persons of both sexes is equal to the number of all possible committees minus the sum of the choices of single-sex committees. All possible committees are C(13,4) = 715 Committees consisting of women only are C(7,4) = 35 Committees consisting of men only are C(6,4) = C(6,2) = 15 715 – 35 – 15 = 665 8. In how many was can we select a committee of four persons so that Mabel and Ralph do not server together? The number of committees where Mabel and Ralph do not server together are equal to the number of all possible committees minus the number of committees where they serve together. All possible committees are 715 For the committees where they serve together, we need to select two persons out of 11, so their number is C(11,2) = 11! / (9! * 2!) = 10*11 /2 = 55 Thus the number of committees where Mabel and Ralph do not server together is 715 – 55 = 660 3 9. In how many ways can we select a committee of four Republicans, three Democrats and two Independents from a group of 10 Republicans, 12 Democrats, and four Independents? C(10,4) * C(12,3) * C(4,2) 10. How many eight-bit strings of 0's and 1's contain exactly three 0's? Here we choose the positions of the 0’s in a string of length 8: C(8,3) = 8! / (3! * 5!) = 6*7*8 / 6 = 56 11. How may eight-bit strings of 0's and 1's contain three 0's in a row? Here we choose the starting position of the three 0’s: 8 – 3 + 1 = 6 In exercises 12-20 find the number of 5 card hands from a standard deck of 52 cards that satisfy the given conditions. 12. Containing all spades The order of the cards in the hand does not matter, and there are 13 spades, so we have C(13,5) 13. All of the same suit Given one particular suit, we C(13,5) choices. There are 4 suits, therefore the choices are 4*C(13,5) 14. Containing cards of exactly two suits Given 2 suits, we have the following cases: suit1 suit2 Choices 1 4 C(13,1) * C(13,4) 2 3 C(13,2) * C(13,3) 3 2 C(13,3) * C(13,2) 4 1 C(13,4) * C(13,1) We can choose the two suits out of 4 suits in C(4,2) ways. Thus the total number of choices is: C(4,2) * (2* C(13,1) * C(13,4) + 2* (C(13,2) * C(13,3))) 4 15. Containing cards of all suits There are 4 suits. In the hand, there will be 2 cards of the same suit and 3 cards from the other three suits. First we choose the suit that will have 2 cards in the hand. The choices are C(4,1) = 1 Next we choose 2 cards from that suit. The choices are C(13,2) Finally we choose three cards of the remaining suits. The choices are independent: 13 x 13 x13 The total number of choices is C(4,1) * C(13,2) * 133 = C(13,2) *2197 = 4*2197 * 13! / (11! * 2!) = 4* 2197 * 13 *12 / 2 = 2107 *13 *6 = 685464 16. Containing an Ace, 2, 3, 4 and 5 all of the same suit. Since the cards are fixed, the only choice is for the suit. There are 4 suits, so the choices are 4 17. Consecutive and of the same suit. (assuming ace is the lowest card) We can choose 5 consecutive cards in a given suit in 13 – 5 + 1 = 9 ways There are 4 suits, thus the total number of choices is 4 * 9 = 36 18. Consecutive (assuming ace is the lowest) For a fixed suit of each card, the choices of 5 consecutive cards are 13 – 5 + 1 = 9 Each of the 5 cards can be of any of the 4 suits, thus we have 45 * 9 = 1024 * 9 = 9216 19. Containing three of one kind and two of another kind. (full house) The choices of the two kinds are permutations of 2 kinds out of 13: P(13,2) = 13*12 = 165 There are four cards of each kind. Thus for each choice of the 2 kinds we have choices that are combinations C(4,2) for the 2 cards and C(4,3) for the 3 cards. Thus the total number of choices is P(13,2) * C(4,2) * C(4,3) 20. Containing two of one kind, two of a second kind and one of a third kind. Consider first choosing the two kinds with 2 cards of each kind. Since the order of the kinds does not matter, the choices are C(13,2). The fifth card can be of any of the remaining 11 kinds. Thus for the choices of the kinds we have C(13,2)*11 2 cards of same kind can be chosen in C(4,2) ways, and one card can be chosen in 4 ways. 5 Thus the total number of choices is 11*C(13,2)*C(4,2)*C(4,2) *C(4,1) = 11*(13*6)*6*6*4 = 20592 In exercises 21-24 a coin is flipped 10 times. 21. How many outcomes are possible? 210 22. How many outcomes have exactly three heads? We choose the position of the heads in the string C(10,3) 23. How many outcomes have at most three heads? Outcomes with 0 heads: Outcomes with 1 head: Outcomes with 2 heads: Outcomes with 3 heads: C(10,0) = 1 C(10,1) = 10 C(10,2) = 45 C(10,3) = 120 The sum is 176 24. How many outcomes have a head on the fifth toss? 29 25. A palindrome is a string that reads the same backwards as forwards. Consider palindromes of digits only (decimal digits 0, 1, 2, …, 9). How many different palindromes are there of length 6? Of length 7? The different palindromes of length 6 are 103 The second half of the string is the same as the first half written in reverse order, so once we choose the first three digits, the string is determined. Each digit is chosen independently of the other digits. Thus we have Cartesian products 10 x 10 x 10 = 1000. The different palindromes of length 7 are 104 The last three digits are the same as the first three digits written in reverse order. We choose the first three digits and the middle digit, i.e. the first 4 digits. Each digit is chosen independently of the other digits. Thus we have Cartesian products 10 x 10 x 10 x 10 = 10,000 6 26. How many 4-digit numbers less than 6000 (decimal number system) can be built using only odd digits? Assume that numbers do not start with leading zeros. Let the number be D1D2D3D4, where Di is the corresponding digit. a. 4-digit numbers D1 can be one of 1, 3, 5. D2, D3 and D4 can be one of 1, 3, 5, 7, 9. Thus all possible 4-digit numbers less that 6000 with odd digits are 3 x 5 x 5 x 5 = 375 b. 3-digit numbers D2, D3, and D4 can be one of 1, 3, 5, 7, 9. Thus all possible 3-digit numbers less that 6000 with odd digits are 5 x 5 x 5 = 125 c. 2-digit numbers D3, and D4 can be one of 1, 3, 5, 7, 9. Thus all possible 2-digit numbers less that 6000 with odd digits are 5 x 5 = 25 d. 1-digit numbers D4 can be one of 1, 3, 5, 7, 9. Here we have 5 numbers. Total: 375 + 125 + 25 + 5 = 530 27. How many 3-digit hexadecimal numbers are there less than F00, but greater than 999? How many of them end in 1? Let the number be D1D2D3, where Di is the corresponding digit. The possible values of D1 are: A, B, C, D, and E The possible values if D2 and D3 are any digit 0 to F Thus all possible numbers of that type are 5 x 16 x 16 = 1280 Numbers that end in 1: 5 x 16 = 80 7 28. A computer password consists of 8 characters. Allowed characters are: Letters in the English alphabet - lower and upper case, decimal digits 0, 1, .., 9, and the underscore symbol. The first two characters must be lower case letters. a. How many different passwords are possible? Since each character in the password is independently chosen, the total number of passwords is a Cartesian product. The first and the second characters are chosen among 26 lower-case letters. Each of the remaining 6 characters is chosen among 63 characters – 26 lower-case letters, 26 upper-case letter, 10 digits and the underscore symbol. Thus we have: 26 x 26 x 63 x 63 x 63 x 63 x 63 x 63 = 262 x 636 = 42,265,887,493,284 b. How many passwords with no repeated characters are possible? The choices for the first two characters are P(26,2). Once the first two characters are chosen, the remaining 6 characters are permutations of 6 element out of 61 element (61 because we have to exclude the two selected lower-case letters for the first two characters) Thus we have all choices to be P(26,2)*P(61,6) 29. How many 3-letter strings formed using letters from A, B, C, D, or E, do not start with the letter A, if letters may be repeated? The first position is chosen among 4 letters, the remaining positions each is chosen among 5 letters: 4 * 5 * 5 = 100 30. How many 3-letter strings formed using letters from A, B, C, D, or E, are there that begin with A or B, if repetition of letters is not allowed? There are 2 choices for the first position. The remaining two positions are permutations of 2 elements over 4 elements: 2*P(4,2) 8