Section 4.3

• 2. P(7,7)=7!= 5040.
• 4. There are 10 combinations and 60 permutations. Below are listed all the 10 combinations followed by their 5 permutations in parentheses: 123 (132 213 231 312 321) 124 (142 214 241 412 421) 125 (152 215 251 512 521) 134 (143 314 341 413 431) 135 (153 315 351 513 531) 145 (154 415 451 514 541) 234 (243 324 342 423 432) 235 (253 325 352 523 532) 245 (254 425 452 524 542) 345 (354 435 453 534 543)
• 6. a) 5 b) C(5,3)=C(5,2)=5.4/2=10 c) 70 d) 1 e) 1 f)924
• 8. P(5,5)=5!=120
• 10. P(6,6)=6!=720
• 12. C(99,2)=4851
• 14. By exercise 53, there are as many subsets with an odd number of elements as there are with an even number of elements. Since there are 2^10=1024 subsets in all, the answer is 1024/2 = 512. This can be done in a different way: C(10,1)+ C(10,3)+C(10,5)+C(10,7)+C(10,9)=10+120+252+120+10=512.
• 16. a) There are C(10,3) ways to choose the positions for the 0's. This is the only choice to be made, so the answer is 120. b) There are C(10,5) ways to choose the positions for the 5 0's. This is the only choice to be made, so the answer is 252. c) We want the number of bit strings with 7,8,9 or 10 1's. By the same reasoning as above there are C(10,7)+C(10,8)+C(10,9)+C(10,10)=120+45+10+1=176 such strings. d) If a string does not have at least 3 1's then it has 0,1,or 2 1's. There are C(10,0)+C(10,1)+C(10,2)=1+10+45=56 such strings. There are 2^10 =1024 strings in all. Therefore there are 1024-56=968 strings with at least 3 1's.
• 18. a) We are not told to worry about the player's positions, so the answer is C(13,10)=286. b) Same as a), except we now have to worry about the order in which the positions are filled. The answer is P(13,10)=1,037,836,800. c) there is only one way to choose the 10 players without choosing a woman, since there are exactly 10 men. Therefore, using part a), there are 286-1=285 ways to choose the players if at least one of them is woman.
• 20. No. of strings of T's and F's of length 40 with exactly 17 T's is C(40,17) approx. = 8.9 x 10^10.
• 22. a) There are C(16,5) ways to select a committee if there are no restrictions. There are C(9,5) ways to select a committee from just the 9 men. Therefore there are C(16,5)-C(9,5)=4368-126 = 4242 committees with at least 1 woman. b) There are C(9,5) ways to select a committee from the 9 men.There are C(7,5) ways to select a committee from just the 7 men. These 2 possibilities do not overlap, since there are no ways to select a committee with neither men nor women. Therefore there are C(16,5)-C(9,5)-C(7,5)=4368-126-21=4221 committees with at least 1 woman and at least 1 man.
• 24. Subtract from the no. of strings with no restrictios the no. of strings that do not contain the letter a. The answer is 26^5 - 25^5=2,115,751. b) If our string is to contain both of these letters, then we need to subtract from the total no. of strings the no. that contain 1 or the other or both of these letters. As in a), 25^5 strings do not contain a and 25^5 strings do not contain b. This is an overcount, since 24^5 strings fail to contain both of these letters. The answer is therefore 26^5-(25^5+25^5-24^5)=312,750. c) First choose the postion for the a; this can be done in 5 ways, since b must follow it. There are 4 remaining positions and these can be filled in P(24,4), since there are 24 letters left (no repetitions allowed). Therefore the answer is 5P(24,4)=1,275,120. d) First choose the positions for the a and b; this can be done in C(5,2) ways, since once we pick 2 positions, we put the a in the left-most and the b in the other. There are 4 remaining positions, and these can be filled in P(24,4) ways, since there are 24 letters left (no repetitions allowed). The answer is C(5,2)P(24,4)=2,550,240.
• 26. We can break down the problem into 3 cases, by sex. There are C(15,6) ways to choose the committee to be composed only of women, C(15,5)C(10,1) ways if there are to be 5 women and 1 man, and C(15,4)C(10,2) ways if there are to be 4 women and 2 men. The answer is the sum of these: 5005+30030+61425=96,460.
• 28. Glue 2 1's to the right of each 0, giving us a collection of 9 tokens: 5 011's and 4 1's. To get the no. of strings consisting of these tokens, all that is involved is choosing the positions for the 1's among the 9 positions in the string, so the answer is C(9,4)=126.
• 30. C(45,3).C(57,4).C(69,5) approx. = 6.3 x 10^16.
• 32. Assume the 1st person sits in the northernmost seat. There are P(5,5) ways to sest the remaining people, since they form a permutation reading clockwise from the 1st person. Therefore the answer is 5!=120.
• 34. We know that C(p,k)=p!/(k!(p-k)!). Clearly, p divides the numerator. On the other hand, p cannot divide the denominator, since the prime factorizations of these factorials contains only the nos. less than p. Therfore the factor p does not cancel when this fraction is reduced to lowest terms (i.e., to a whole no.), so p divides C(p,k).
• 36. C(13,8)=1287.
• 38. C(11,7)1^4=330.
• 40. C917,9).3^8.2^9=81,662,929,920.
• 44. We add adjacent nos. in this row to obtain the next row, starting and ending with 1: 1 11 55 165 330 462 330 165 55 11 1
• 56. By exercise 55, there are C(n-k+k,k)=C(n,k) paths from (0,0) to (n-k,k) and C(k+n-k,n-k)=C(n,n-k) paths from (0,0) to (k,n-k). By symmetry, these 2 quantities must be the same (flip the picture around the 45 degree line).