Discrete Mathematics- An Open Introduction - 3rd Edition, 2016a

354 B. Selected Solutions
1.7.6. 51 passengers. We are asking for the size of the union of three
non-disjoint sets. Using PIE, we have 25 + 30 + 20 − 10 − 12 − 7 + 5 51.
1.7.7.
(a) 2 8 strings.
(b) ( 8
5) strings.
(c) ( 8
5) strings.
(d) There is a bijection between subsets and bit strings: a 1 means that
element in is the subset, a 0 means that element is not in the subset.
To get a subset of an 8 element set we have an 8-bit string. To make
sure the subset contains exactly 5 elements, there must be 5 1’s, so
the weight must be 5.
(
1.7.8. 13
(
10)
+
17
)
8 .
1.7.9. With repeated letters allowed, we select which 5 of the 8 letters will
be vowels, then pick one of the 5 vowels for each spot, and finally pick one
of the other 21 letters for each of the remaining 3 spots. Thus, ( 8) 5 5 5 21 3
words.
Without repeats, we still pick the positions of the vowels, but now
each time we place one there, there is one fewer choice for the next one.
Similarly, we cannot repeat the consonants. We get ( 8
5) 5!P(21, 3) words.
1.7.10.
(a) ( 5) ( 11
2 6
) paths.
(b) ( 16) (
8 − 12
) ( 4
)
7 1 paths.
(c) ( 5) ( 11
) (
2 6 + 12
) ( 4
) (
5 3 − 5
) ( 7
) ( 4
)
2 3 3 paths.
(
1.7.11. 18
) ( ( 18
) )
8 8 − 1
exactly one fewer route back.
1.7.12. 2 7 + 2 7 − 2 4 strings (using PIE).
routes. For each of the ( 18) 8 routes to work there is
1.7.13.
( 7
3
) ( + 7
) (
4 − 4
)
1 strings.
1.7.14. There are 4 spots to start the word, and then there are 3! ways to
arrange the other letters in the remaining three spots. Thus the number of
words avoiding the sub-word “bad” in consecutive letters is 6! − 4 · 3!.
If we now need to avoid words that put “b” before “a” before “d”, we
must choose which spots those letters go (in that order) and then arrange
the remaining three letters. Thus, 6! − ( 6
3) 3! words.
1.7.15. 2 n is the number of lattice paths which have length n, since for
each step you can go up or right. Such a path would end along the line