
Chapter 4
 #20: Let K represent the total number of letters
in the word, and use L_{i} to represent
the number of repetitions of each letter, 1
<= i <= 3, i an integer, such that L_{1}
+ L_{2} + L_{3} = K. We need,
then, all unique arrangements of the letters such
that K!/(L1!L2!L3!) = (L_{1} + L_{2}
+ L_{3})!/(L_{1}!L_{2}!L_{3}!)
<= 1000.
 #22: We assume here that each of the 26
letters occurs once. The letters c and d represent
two of the 26 letters. There are 2! = 2 ways to
arrange these two letters. Now we select 6 of the
remaining 24 letters and permute them between c
and d. This can be done in C(24,6)×P(6,6) =
P(24,6) ways. This gives us a bundle of 6 letters
(none to be c or d) surrounded by c and d. Now we
permute the remaining 18 letters in P(18,18) = 18!
ways. Within any arrangement of these 18 letters,
there are 19 slots into which we can place the
8letter bundle described previously. We choose
one of those 19 slots, in C(19,1) = 19 ways. In
all, then, there are P(2,2)×P(24,6)×P(18,18)×19
desired arrangements. Another way to consider this
is to first permute the 24 letters no including c
or d. This can be done in P(24,24) = 24! ways.
Now, we count the number of slotpairs we can
place c and d such that they surround exactly 6
letters. In the line up of 24 letters, there are
25 slots. Let us call those slots 0 through 24. A
slotpair must have 6 letters between them, so the
pair 0/6 is the first available slotpair,
continuing through slotpair 1/7, 2/8, and so on,
through 18/24. There are 19 different slotpairs
to choose from in placing the letterpair c/d. We
can use d/c as well. Therefor, there are 24!×19×2
total unique arrangements. The two expressions we
have generated can be shown to be equivalent.
Chapter 5 Section 1
 #1: One way to solve this problem is to create a
threecircle overlapping Venn diagram and start
from the deepest overlap (all three magazines) and
work your way out. Remember to include the 8
students who subscribe to none of these magazines.
Now count all eight components of your diagram.
You should get 99 in all. Another way to solve
this problem is to solve use the
Inclusion/Exclusion principle and solve for T, the
total number of elements in a set. Here, we first
generate the I/E equation 8 = T  (47 + 44 + 32) +
(11 + 12 + 12)  3. Solving for T we get T = 99.
 #2: (a) There are 2^{4} = 16 different
descriptions possible for a diner, because each
diner is characterized using 4 categories, and
each diner either has or does not have each
categorical characterization. (b) We seek the
number of nonnegative integer solutions to the
equation c_{1} + c_{2} + . . . + c_{16}
= 38. We have 16 categories and 38 objects to
place within them, so there are C(16  1 + 38, 16
1) = C (16  1 + 38, 38) different ways to fill
the categories with the 38 objects.
 #4: Use either the I/E principle or a Venn
diagram to generate data that sets up a
contradiction. Using the I/E principle, we have
the equation 111 = 144  (16 + 13 + 17) + (10 + 9
+ X)  5. Solving for X, which represents the
number of people who had been to both Europe and
Asia, we get X = 1, which is impossible.
 #7: By the I/E principle, we solve the
following inequality for T: 1 <= T  (15 + 14 +
18) + (8 7 + 9)  6. This yields 30 <= T.
So, the smallest total enrollment is 30.

