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Unformatted text preview: as 136 E Chapter 5 Counting 36. Let K be the number of other peeple at the party that person :1: knows. The possible values for K are 0, 1, . . . ,n  1, where n 3 2 is the number of people at the party. We cannot apply the pigeonhole principle directly, since there are n pigeons and n. pigeonholes. However, it is impossible for both 0 and n. — 1 to be in the range of K, since if one person knows everybody else, then nobody can know no one else (we assume that “knowing” is symmetric). Therefore the range of K has at most 11. — 1 elements, whereas the domain has n elements, so K is not onetoone, precisely what we wanted to prove. 38. a) The solution of Exercise 37, with 24 replaced by 2 and 149 replaced by 127, tells us that the statement is true. a b) The solution of Exercise 37, with 24 replaced by 23 and 149 replaced by 148, tells us that the statement is true. c) We begin in a manner similar to the solution of Exercise 37. Look at r11, (12, . . . , 1175, :21 +25, . . . , 1175 +25, _ where o, is the total number of matches played up through and including hour 2'. Then 1 3 (11 < {1.2 < it  i < 11.75 g 125, and 26 g 0.1 + 25 < (1.2 + 25 <   < 275 + 25 g 150. Now either these 150 numbers are precisely all the number from 1 to 150, or else by the pigeonhole principle we get, as in Exercise 37, o, = a, + 25 for
some 2' and j and we are done. In the former case, however, since each of the numbers 11., + 25 is greater than or equal to 26, the numbers 1,2,. . . , 25 must all appear among the (his. But since the o,’s are increasing, the only way this can happen is if 411 = 1 , a2 = 2, . . . , {1.25 = 25. Thus there were exactly 25 matches in the ﬁrst 25 hours.
d) We need a different approach for this part, an approach, incidentally, that works formany numbers besides 30 in this setting. Let 0.1, (32, ..., (1.75 be as before, and note that 1 g 0.1 < (12 < < 1275 g 125.. By the pigeonhole principle two of the numbers among a1, :12, . . . , 1231 are congruent modulo 30. If they differ by 30, then we have our solution. Otherwise they differ by 60 or more, so 1231 Z 61. Similarly, among (131
through 4251, either we ﬁnd a solution, or two numbers must differ by 60 or more; therefore we can assume that (151 Z 121. But this means that 1266 Z 126, a contradiction. 40. Look at the pigeonholes {1000,1001}, {1002,1003}, {1004,1005}, ..., {1098,1099}. There are clearly 50
sets in this list. By the pigeonhole principle, if we have 51 numbers in the range from 1000 to 1099 inclusive, then at least two of them must come from the same set. These are the desired two consecutive house numbers. 42. Suppose this statement were not true. Then for each 2', the 3th box Contains at most a, — 1 objects. Adding, +
we have at most (111 — 1) + (11.3 —— 1) +  r  + (n, — 1) = m + 71.2 + i   + m — 75 objects in all, contradiCting the
fact that there were 121 + 71.2 +    + at — t + 1 objects in all. Therefore the statement must be true. SECTION 5 .3 Permutations and Combinations
2. P(7, 7) = 7! = 5040 4. There are 10 combinations and 60 permutations. We list them in the following way. Each combination is listed, without punctuation, in increasing order, followed by the ﬁve other permutations involving the same numbers, in parentheses, without punctuation. _ h
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) Section 5.3 Permutations and Combinations 137 a. a) C(5,1)=5 b) C(5,3)=C(5,2)=54/2=10 c) C(8,4)=876'5/(432)=70 d) C(8,8)=1 e) C(8,0)=1" f) C(12,6)=121110987/(65432)=924
s. P(5,5)=5!=120 10. P(6,6) = 6! = 720 12. a) To specify a bit string of length 12 that contains exactlythree 1‘s, we simply need to choose the three positions that contain the 1’s. There are C (12, 3) = 220 ways to do that.
b) To contain at most three 1’s means to contain three 1’s, two 1’s, one 1, or no 1’s. Reasoning as in' part (a), we see that there are C(12, 3) +C(12, 2) +C(12, 1) +C(12, 0) = 220+66+ 12+ 1 = 299 such strings.
c). To contain at least three 1’s means to contain three 1’s, four 1’s, five 1’s, six 1’s, seven 1’s, eight 1’s,
nine 1’5, 10 1’3, 11 1’3, or 12 1’s. We could reason as in part (b), but we would have too many numbers to add. A simpler approach would be to ﬁgure out the number of ways not to have at least three 1’s (i.e., to
have two 1’s, one 1 , or no 1’s) and then subtract that from 212, the total number of bit strings of length 12. This way we get 4096 — (66 + 12 + 1) = 4017.. .
d) To have an equal number of 0’s and 1’s in this case means to have six 1’s. Therefore the answer is C(12, 6) = 924.
14. C(99, 2) = 99  98/2 = 4351 16. We need to compute C(10, 1) + C(10, 3) + C(10, 5) + C(10, 7) + C(10, 9) = 10 + 120 + 252 + 120 +10 = 512.
(In the next section we will see that there are just as many subsets with an odd number of elements as there are subsets with an even number of elements (Exercise 31 in Section 5.4). Since there are 210 = 1024 subsets in all, the answer is 1024 / 2 = 512, in agreement with our computation.) 18. a) Each ﬂip can be either heads or tails, so there are 28 = 256 possible outcomes.
b) To specify an outcome that has exactly three heads, we simply need to choose the three ﬂips that came up heads. There are C’ (8, 3) == 56 such outcomes. . _
c) To contain at least three heads means to contain three heads, four heads, ﬁve heads, six heads, seven heads, or eight heads. Reasoning as in part (b), we see that there are C(8, 3) + C(8, 4) + C(S, 5) + C(8, 6) + C(8, 7) +
:C (8, 8) = 56 + 70 + 56 + 28 + 8 + 1 = 219 such outcomes. We could also subtract from 256 the number of
 ways to get two of fewer heads, namely 28 + 8 + 1 = 37., Since 256 —— 37 = 219, we obtain the same answer ' using this alternative method. I
d) To have an equal number of heads and tails in this case means to have four heads. Therefore the answer is C(8,4) = 70. 20. a) There are C (10, 3) ways to choose the positions for the 0’s, and that is the only choice to be made, so the answer is C(10, 3) r: 120. r
b) There are more 0’s than 1’5 if there are fewer than ﬁve 1’s. Using the same reasoning as in part (a),
together with the sum rule, we obtain the answer C(10,0) + C(10, 1) + C(10,2) + C(10,3) + C(10, 4) =
1 + 10 + 45 + 120 + 210 = 386.. Alternatively, by symmetry, half of all cases in which there are not ﬁve 0’s
have more 0’s than 1’8; therefore the answer is (210 ~— 0 (10, 5) / 2 = (1024 — 252) / 2 = 386. 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 l 10 + 1 = 176 such strings. 
d) If a string does not have at least three 1’s, then it has 0, 1, or 2 1’5. There are C(10,0) + C(10, 1) +
C(10, 2) = 1 + 10 +45 = 56 such strings. There are 210 = 1024 strings in all. Therefore there are 1024 — 56 = 968 strings with at least three 1’s. 133 Chapter 5 Counting 22. a) If ED is to be a substring, then we can think of that block of letters as one superletter, and the problem
is to count permutations of seven items—the letters A, B, C, F, G, and H, and the superletter ED. Therefore the answer is P(7, 7) = 7! r: 5040.
_b) Reasoning as in part (a), we see that the answer is P(6, 6) = 6! = 720.
c) As in part (a), we glue BA into one item and glue FGH into one item. Therefore we need to permute ﬁve items, and there are P(5, 5) = 5! = 120 ways to do it.
d) This is similar to part Glue AB into one item, glue DE into one item, and glue GH into one item, producing ﬁve items, so the answer is P(5, 5) = 5! = 120.
e) If both CAB and BED are substrings, then CABED has to be a substring. So we are really just permuting four items: CABED, F, G', and H. Therefore the answer is P(4, 4) = 4! = 24. There are no permutations with both of these substrings, since B cannot be followed by both C and F at ' the same time. 24. First position the women relative to each other. Since there are 10 women, there are P(10, 10) ways to
do this. This creates 11 slots where a man (but not more than one man) may stand: “in front of the ﬁrst woman, between the ﬁrst and second women, . . . , between the ninth and tenth women, and behind the tenth
woman. We need to choose six of these positions, in order, for the ﬁrst through six man to occupy (order matters, because the men are distinct people). This can be done is P(11, 6) ways. Therefore the answer is
P(10, 10)  P(11, 6) = 10!  11!/5! == 1,207,084,032,000. 26. a) This is just a matter of choosing 10 players from the group of 13, since we are not told to worry about what positions they play; therefore the answer is C (13, 10) :286.
b) This is the same as part (3), except that we need to worry about the order in which the choices are made,
since there are 10 distinct positions to be ﬁlled. Therefore the answer is P(13, 10) = 13! / 3! :2 1,037,836,800. c) There is only one way to choose the 10 players without choosing a woman, since there arelexactly 10 men.
Therefore (using part (21)) there are 286 —— 1 :2 285 ways to choose the players if at least one of them must be EL woman. 28. We are just being asked for the number of strings of TS and F’s of length 40 with exactly 17 T’s. The only
choice is which 17 of the 40 positions are to have the T’s, so the answer is C(40, 17) W 8.9 x 1010. 30. 3.) There are C(16, 5) ways to select a committee if there are no restrictions. There are CG), 5) ways to select
a committee from just the 9 men. Thereforethere are C(16, 5) ~— 0 (9, 5) = 4368 — 126 = 4242 committees with at least one woman.
b) 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. There are C(7, 5) ways to select a committee from just the 7 men. These two possibilities do not overlap, since there are no ways to select a committee containing neither men nor
women. Therefore there are C(16, 5) # 0(5), 5) —— C (7, 5) = 4368 —— 126 — 21 = 4221 committees with at least one woman and at least one man. 32. a) The only reasonable way to do this is by subtracting from the number of strings with no restrictions the number of strings that do not contain the letter a. The answer is 266 — 256 = 308915776 — 244140625 = 64,775,151.
b) If our string is to contain both of these letters, then we need to subtract from the total number of strings the number that fail to contain one or the other (or both) of these letters. As in part (a) , 256 strings fail to contain
an a; similarly 256 fail to contain a b. This is overcounting, however, since 246 fail to contain both of these letters. Therefore there are 256 — 256 ~— 245 strings that fail to contain at least one of these letters. Therefore
the answer is 265 — (256 + 256 — 245) =' 308915776 — (244140625 + 244140625 —*~ 191102976) 1:: 11,737,502. 139 I) must follow it. There are four in 5 ways, since the
24 letters left (no repetitions c) First choose the position for the a; this can be done
remaining positions, and these can be ﬁlled in 13(24, 4) ways, since there are being allowed this time). Therefore the answer is 5P(24, 4) = 1,275,120. (1) First choose the positions for the a. and b; this can he done in C(6,2) ways, since once we pick two
'ning positions, and these positions, we put the a in the left—most and the b in the other. There are four rema1 can be ﬁlled in P(24, 4) ways, since there are 24 letters left ( and one man, and C(15), 4)C(10, 2) ways if there are to be
C(15, 6) + C(15, 50000, 1) + C(15, 4)C(1 right of each 0, giving'us a collection of nine tokens: ﬁve 011‘s and four 1’s. We are
5. All that is involved is choosing the positions for 33. C(45,3)  C(57,4)  C(69, 5) = 14190.39501011238513 e 6.3 x 1016 erson sits in the northernmost seat. Then there are P (5, 5) ways to 40. We might as well assume that the ﬁrst p
utation reading clockwise from the ﬁrst person. Therefore seat the remaining people, since they form a perm "'_'.;'.'." ' 4'
FIElg. 331 I. the answer is 5!: 120. I .
.7... 'nteger between 0 and 5 , inclusive. The scoring ' . There are C(5,p) ways to choose
of p from 0 to 5. 140 ' Chapter 5 Counting The sum is 252. So there are 252 ways for the score to end up tied. We already noted in the paragraph above
_ that there are 672 different scoring scenarios, so there are 672 — .252 = 420 scenarios in which the score is not ' tied. This answers the question for this part of the exercise. _ b) This is easy after what we‘ve found above. There are 252 ways for the score to be tied at the end of the ﬁrst group of penalty kicks, and there are 420 ways for the game to be settled in the second group. SO there
are 252  420 = 105,840 ways for the game to end during the second round. I I c) We have already seen that there are 420 ways for the game to end in the ﬁrst round, and 105,840 more ; ways for it to end in the second round. In order for it to go into a sudden death period, the ﬁrst two rounds
must have ended tied, which can happen in 420  420 =2 176,400 ways. Thereafter, the game can end after two more kicks in 2 ways (either team can make their kick and have the other team miss theirs), after four more
kicks in 2  2 = 4 ways (the ﬁrst pair of kicks must have the same result, either both made or both missed,
and then either team can win), after six more kicks in 22  = 8 ways (the ﬁrst two pairs of kicks must have
the same results, and then either team can win), after eight more kicks in 16 ways, and after ten more kicks
in 32 ways. Thus there are 2 + 4 + 8 + 16 + 32 = 62 ways for the sudden death round to end within ten kicks.
This needs to be multiplied by the 176,400 ways we can reach sudden death, for a total of 10,936,800 scoring scenarios. So the answer to this last question is 420 + 105840 + 10936800 2 11,043,060. added. Terms of the form :35, 3:41;, 3:3y2, s2y3, spit and 315 arise. To obtain a term of the form a: , an :1: must be chosen in each of the sums, and this can be done in only one way. Thus, the :85 term in the product
has a coefﬁcient of 1. (We can think of this coefﬁcient as To obtain a term of the form 3343;, an a: must
be chosen in four of the ﬁve sums (and consequently a. y in the other sum). Hence, the number of such terms is the number of 4combinations of ﬁve objects, namely = 5. Similarly, the number of terms of the form 2:33;? is the number of ways to pick three of the ﬁve sums to obtain :r’s (and consequently take a y from each
5 of the other two factors). This can be done in 2 10 ways. By the same reasoning there are (2) = 10 ways to obtain the s2y3 terms, == 5 ways to obtain the my"él terms, and only one way (which we can think of as
5 to obtain a 5 term. Consequently, the product is 335 + 51:43; + 10:1."3y2 + 10:1:2y3 + 5:5y4 + 3/5.
0 y
s .... b) This is explained in Example 2. The expansion is $5 + 34y + 2231,? + $2343 + 2:394 + — _
:175 + 53343; + 10:33:!)2 + 10.732393 + 5:1:y4 + y5. Note that it does not matter whether we think of the bottom of the
binomial coefﬁcient expression as corresponding to the exponent on 3:, as we did in part (a), or the exponent on y, as we do here. 4. (133) = 1287 6. (1,1)14 = 330 s. (1,7) 3329 = 243106561  512 = $662,929,920 10. By the Binomial Theorem, the typical term in this expansion is (120)3100'j (1 / 3:)j , which can be rewritten as (190)s100"2j. As 3' runs from 0 to 100, the exponent runs from 100 down to —~100 in decrements of 2. If we 3
let it denote the exponent, then solving k = 100— 29‘ for 3' we obtain 3' = (100 — k) / 2. Thus the values of k
100 and for such values of k for which ask appears in this expansion are —100, —98, ..., —2, 0, 2, 4, ..., , the coefﬁcient is ((10331) /2) . Section 5.4 Binomial Coefﬁcients 141 ' 12. We just add adjacent numbers in this row to obtain the next row (starting and ending with 1 , of course): 1 , 1 11 55' 165 330 462 462 330 165 55 11 1 14. Using the factorial formulae for computing binomial coefﬁcients, we See that ($1) k (n)  If k < “Tl/2: k < 1, so the “less than” signs are correct. Similarly, if k > 11/2, then 11%“ then n—k+1 than” signs are correct. The middle equality is Corollary 2 in Section 5.3, since [11/2] + i equalities at the ends are clear. n —— 1 binomial coefﬁcients through (nil). 16. a) By Exercise 14, we know that (WT/‘21) is the largest of the
. But since 211. g 2'” for n 2 2, it Therefore it is at least as large as their average, which is (2” —— 2) /(n, — 1)
fellows that an  2l/(n  1) Z 271/11,, and the proof is complete. b) This follows from part (a) by replacing n with 271. when n 2 2, these coefﬁcients are single digit numbers in base I
the numeral formed by concatenating the symbols in the fourth row of Pascal‘s triangle is the answer. (11. — 1)ln!(n + 1)!
(k — 1)!kl(k + 1)!(n. — k —— 1)l(n — k)l(n — k + 1)! ' set with n elements, and we wish to choose a subset A with 1:: elements and
The lefthand side gives us the number of ways to do this, #' lements that are to go into one or the other of the n r _ n! r' __ (7.) (k) " 7K.” __ 7.)! kw,» _ k)! _ him, —~ 'r)l('r —— k)! 1 and on the other hand n n — k __ n‘ (71 — kl' _ n1 _ k) m. _ a We — mm  w “ rm — m — a 24. We know that
k — kl(p — k)! ' Clearly p divides the numerator. On the other hand, p cannot divide the denominator, since the prime
tor p does not cancel factorizations of these factorials contains only numbers less than 33. Therefore the fac
d to lowest terms (i.e., to a whole number), so 39 divides . chairman from among the n mathematicians (
e from among the other 291. —— 1 professors. This gives us 71 members of the committe on the right—hand side. On the other hand, for each k: from 1 to n, we can
mathematicians and n. — k: computer scientists. There are (k to choose the chairman from among these, and (milk
, we obtain the expression on theleft—hand side of the identity. quantity equals 34. By Exercise 33 there are (“'2”) = (0, 0) to (km — It). By symmetry, these two quantities must be line) . 36. A path ending up at, (n + 1 ~— k,k) 11, ways) times the number of ways to choose the other 72. — 1 2n—1 
( 1"#1), the expressmn have our committee consist of k
11') ways to choose the mathematicians, )9 ways ) ways to choose the computer scientists. Since this last paths from (0,0) to (n  k,k)' and ( 1%,; must have made its last step either upward or to
(n + 1 — k, k — 1); if it was made to the right, then it came from
number of paths to ...
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 Chairman, ways, Polycephaly, Conjoined twins, n. pigeonholes, CABED

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