Transcript Discrete Mathematics Math 6A

Discrete Mathematics
Math 6A
Homework 3 Solution
4.1-3 (a) 4^10 (b) 5^10
4.1-16 26^4 – 25^4 = 66,315
4.1-21 (a) 1000-10 = 990 (b) 5*10*10=500 (c) 3*9=27
4.2-2 This follows from the pigeonhole principle, with k=26
4.2-3 (a) 3 (b) 14
4.2-14 (a) We can group the first ten positive integers into five subsets of two
integers each, each subset adding up to 11: {1,10},{2,9},{3,8},{4,7} and {5,6}. If
we select seven integers from this set, then by the pigeonhole principle at least
two of them come from the same subset. Furthermore, if we forget about these
two in the same group, then there are five more integers and four groups; again
the pigeonhole principle guarantees two integers in the same group. This gives
us two pairs of integers, each pair from the same group. In each case these two
integers have a sum of 11, as desired.
(b) No. The set {1,2,3,4,5,6} has only 5 and 6 from the same group, so the only pair
with sum 11 is 5 and 6.
4.2-19 (a) If this statement were not true, then there would be at most 8 from each
class standing, for a total of at most 24 students. This contradicts the fact that there are
25 students in the class
(b) If this statement were not true, then there would be at most 2 freshmen, at most 18
sophomores, and at most 4 juniors, for a total of at most 24 students. This is
contradicts the fact that there are 25 students in the class
4.2-40 Look at the pigeonholes {1000,1001},{1002,1003}, {1004,1005},...,{1098,1099}.
There are clearly 50 sets in this list. So 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.
4.3-4 a) 60 permutations: P(5,3)
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)
4.3-9 P(12,3) = 1320
4.3-11
(a) C(10,4)=210
(b) C(10,4) + C(10,3) + C(10,2) + C(10,1) + C(10,0) = 386
(c) C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10) = 848
(d) C(10,5)= 252
4.3-18
(a) 2^8 = 256
(b) C(8,3) = 56
(c) C(8,3) + C(8,4) + C(8,5) + C(8,6) + C(8,7) + C(8,8) = 219
(d) C(8,4) = 70
4.3-40 5! = 6!/6 = 120
4.4-2
(a) (x+y)^5 = (x+y) (x+y) (x+y) (x+y) (x+y)
= x^5+5(x^4)y +10(x^3)(y^2) + 10(x^2)(y^3) + 5x(y^4) + y^5
(a) C(5,0)x^5 + C(5,1)(x^4)y + C(5,2)(x^3)(y^2) + C(5,3)(x^2)(y^3) + C(5,4)x(y^4) +
C(5,5) y^5 = x^5+5(x^4)y +10(x^3)(y^2) + 10(x^2)(y^3) + 5x(y^4) + y^5
4.4-9 –(2^10)(3^99)C(200,99)
4.4-21
a) We show that each side counts the number of ways to choose from a set with n
elements a subset with k elements and a distinguished element of that set. For
the left-hand side, first choose the k-set (C(n,k) ways) and then choose one of
the k elements in this subset to be the distinguished element (k ways). For the
right-hand side, first choose the distinguished element out of the entire n-set n-1
elements of the set (C(n-1,k-1) ways)
b) This is straightforward algebra:
kC(n,k) = k{n!/(k!(n-k)!)} = (n(n-1)!)/((k-1)!(n-k)!) = nC(n-1,k-1)
4.4-33
a) Clearly a path of the desired type must consist of m moves to the riht and n
moves up. Therefore each such path can be represented by a bit string
consisting of m 0's and n 1's, with the 0's representing moves to the right and the
1's representing moves up. Note that the total length of this bit string is m+n
b) We know from this section that the number of bit string of length m+n containing
exactly n 1's is C(m+n, n), since one need only specify the positions of the 1's.
Note that this is the same as C(m+n, m)
4.4-34 By 33 there are C(n-k+k, k)=C(n,k) path from (0,0) to (n-k,k) and C(k+n-k, nk)=C(n,n-k) path from (0,0) to (k, n-k). By symmetry, these two quantities must be
the same