
Chapter 3 Section
2
 #3: P(14,4) as (a) product of 4
consecutive integers: 14 × 13 × 12 × 11; (b) using
factorial notation: 14!/(10!); (c) in terms of
C(14,4): C(14,4)*4!
 #5: C(42,19) = C(42,4219) = C(42,23),
so r = 23
 #11: (a) There are no restrictions about
type of token nor are we concerned about order the
token are chosen, so we pool the tokens and choose
20: C(68,20). (b) Now we only select from those of
the same color. There are C(26,20) ways to select
20 token all red and C(42,20) to select 20 green
tokens. Because these are disjoint selections and
we are choosing one token, we add these results.
This yields C(26,20) + C(42,20) ways to select 20
tokens all of one color.
 #15: There are P(14,14) = 14! ways to
arrange the 14 letters in the word, but some of
those arrangements will be duplicates, because we
have some letters appearing more than once. For
instance, if we label the two L letters as L_{1}
and L_{2}, we have these two arrangements
among the 14! that are possible: QUADRIL_{1}L_{2}IONTHS
and QUADRIL_{2}L_{1}IONTHS.
Because the letters L are identical, these two
arrangements are the same. To account for this
duplication, we divide 14! by the number of ways
each of the duplicated letters can be permuted.
This gives u 14!/(2!2!) total unique arrangements.
 #18: We need to have 40 years × 365.25
days/year (leap year inclusion) = 14,610 3joke
sets, with the order the jokes are told not
distinguishing one set of three from another. This
leads to the need to determine the value of n
such that C(n,3) ≥ 14,610. You can
substitute C(n,3) = (n!)/[(n3)!3!]
and solve the resulting cubic inequality or you
can use an educatedguessandcheck strategy to
determine n. Here, the smallest integer
value of n that satisfies the inequality
is n = 46. Therefore, Juanita must have at
least 46 jokes in her joke set.
 #19: (a) We need to consider distinct
cases in order to avoid overcounting. To assure
at least 3 faculty and 3 students, we have three
distinct cases:
 Case I: exactly 3 faculty and 5 students. We
can have C(20,3) * C(20,5) different committees
of this make up.
 Case II: exactly 4 faculty and 4 students. We
can have C(20,4) * C(20,4) different committees
of this make up.
 Case I: exactly 5 faculty and 3 students. We
can have C(20,5) * C(20,3) different committees
of this make up.
 Because each case has no overlap with other
cases, we add to determine the total number of
committees that are possible:
 C(20,3) × C(20,5) + C(20,4) × C(20,4) +
C(20,5) × C(20,3) = 2[C(20,3) × C(20,5)] +
[C(20,4)]^{2}.
Note that C(20,3)
× C(20,3) × C(34,2) is not
a correct solution nor does it illustrate a
correct strategy!
(b) Here, it
is more efficient to determine the total number of
committees possible and remove those with no
faculty: C(40,8)  [C(20,0) × C(20,8)].
Chapter 3 Section 3
 #1: C(7,k), with 0 ≤ k ≤
7
 #7: (a) We need to traverse 8 blocks
southward and 12 blocks eastward. This yields
C(20,8) = C(20,12) 20block paths from T to B. (b)
One way to solve this problem is to consider the
number of possible paths with no restrictions,
C(20,8) = C(20,12), then count and remove those
that can't be traversed. In the scenario described
here, it is a 9block drive to the west end of the
street to be closed. There are C(9,4) = C(9,5)
ways to get to that point. There is only one path
from that point to the east end of that restricted
street: C(1,1). Finally, there are C(10,4) =
C(10,6) paths from that eastend point to point B.
We multiply these three counts together, to match
them: C(9,5) × C(1,1) × C(10,6). We then subtract
this number of closed paths from the original
total number of paths: C(20,12) − [C(9,5) × C(1,1)
× C(10,6)].
 #11: This is analogous to the
birthorder problems we explored: C(6,4) = C(6,2).
 #26: (a) 21!/(3!2!2!2!2!2!2!3!) (b)
There are 11 ways to choose the first (leftmost)
consonant and 10 ways to choose the last
(rightmost) consonant. There remain 19 letters to
permute, to be placed between these two
consonants. We still need to account for all the
duplication of arrangements due to repetition of
letters. This yields
(11×10×19!)/(3!2!2!2!2!2!2!3!) different such
arrangements. (c) Concentrating on the vowels, we
have 10!/(3!3!!2!2!) different arrangements of the
vowels. We now have 11 consonants to arrange,
yielding 11!/(2!2!!2!2!) different consonant
arrangements. Because there are no restrictions on
how the consonants are positioned (relative to
consonants only), we can think of the 11
consonants as borders and the 12 spaces around the
consonants as positions within which the one vowel
group could be placed. This yields 12 spots into
which we could drop the vowel group. We now match
each consonant arrangement with each vowel
arrangement, using the Multiplication Principle,
and, finally, multiply that by 12: 12 ×
10!/(3!3!!2!2!) × 11!/(2!2!!2!2!) different
arrangements. (d) There are 11 unique letters in
the word, and we select 5 of them, with no regard
for order: C(11,5).

