
Chapter 3 Section
4
 #14: (a) There are 5^{10} uncollected
terms. (b) R = (10!)/(3!1!2!4!)
 #20: (w  2k)^{4} = C(4,0)w^{4}(2k)^{0}
+ C(4,1)w^{3}(2k)^{1} + C(4,2)w^{2}(2k)^{2}
+ C(4,3)w^{1}(2k)^{3} + C(4,4)w^{0}(2k)^{4}=
w^{4} + 4w^{3}(2k) + 6w^{2}(2k)^{2}
+ + 4w(2k)^{3} + (2k)^{4} = w^{4}
 8w^{3}k + 24w^{2}k^{2} 
32wk^{3} + 16k^{4}
 #27: (a) 4^{13} uncollected
terms; (b) C = 13!/(2!3!5!3!)
Chapter 4
 #1: We know that the number of
uncollected terms is 5^{15}, so, after
collecting terms, the sum of the coefficients must
still be 5^{15}.
 #4: Arrange the 8 Qs. There's just one way to do
this, assuming the Qs are identical. Now, among
the 8 Qs, there are 9 spots for the Ms such that
the Qs keep the Ms nonadjacent. Choose 4 of these
spots and place the Ms. This can be done in C(9,4)
ways.
 #5: We'll speak in terms of $5 units. There are
20 $5 units in $100. (a) This problem is analogous
to determining the number of positive integer
solutions to the equation x_{1} + x_{2}
+ ... + x_{10} = 20. Because there are 10
categories, a tally sheet would have 9 dividers.
Because there are 20 objects ($5 units) to
separate among the 10 categories, there are 19
places among the 20 objects into which we can
place the 9 dividers. This can be done C(19,9)
ways. (b) The phrase "at most" indicates one or
more of the 10 charities could get nothing. This
version of the problem is analogous to determining
the number of nonnegative integer solutions to
the equation x_{1} + x_{2} + ... +
x_{10} = 20. Because there are 10
categories, a tally sheet would have 9 dividers.
There are 20 objects ($5 units) to distribute
among up to 10 categories. Thus, we have 29
elements to arrange, 9 of which are identical and
the other 20 are identical. There are C(29,9) =
C(29,20) arrangements.
 #8: We are determining the number of
nonnegative integer solutions to the equation z +
y + x + a + b + c = 28. Because there are 6
variables, a tally sheet would have 5 dividers.
There are 28 units to distribute among up to
6variables . Thus, we have 33 elements to arrange,
5 of which are identical and the other 28 are
identical. There are C(33,5) = C(33,28)
arrangements. Thus, there are C(33,5) = C(33,28)
nonnegative integer solutions to the equation z +
y + x + a + b + c = 28. Any example solution that
includes at least one variable equal to 0, and the
six sum to 28, is correct. One such equation is 4
+ 0 + 20 + 1 + 0 + 3 = 28.
 #9: (a) There are 15 elements in the set, 1 of
type A, 2 of type B, 3 of type C, 4 of type D, and
5 of type E. Thus, there are 15!/(1!2!3!4!5!)
distinct arrangements. (b) There are 6 vowels and
9 nonvowels. Use the 9 nonvowels as dividers.
These 9 nonvowels can be arranged in 9!/(2!3!4!)
ways. Now, among the 9 nonvowels, there are 10
spots into which we can place the vowels so that
the vowels will be non adjacent. Choose 6 of these
10 spots, which can be done in C(10,6) ways. Once
the spots for the vowels are selected, we
determine the number of ways to arrange the
vowels. This can be done in 6!/(1!5!) ways. We use
the multiplication principle to determine the
total number of arrangements, because we are
matching the results of each step. This means
there are 9!/(2!3!4!) × C(10,6) × 6!/(1!5!) ways
to arrange the 15 elements of the set so that no
two vowels are adjacent.
 #14: In (a + b + c)^{10}, the
coefficient 10!/(3!5!2!) means that one variable
is raised to the 3rd power, another to the 5th
power, and another to the 2nd power. There are 3!
= 6 ways to arrange the three variables, so there
must be 6 collected terms,
including 10!/(3!5!2!) × a^{3}b^{5}c^{2},
10!/(3!5!2!) × a^{3}c^{5}b^{2},
10!/(3!5!2!) × b^{3}a^{5}c^{2},
10!/(3!5!2!) × b^{3}c^{5}a^{2},
10!/(3!5!2!) × c^{3}a^{5}b^{2},
and 10!/(3!5!2!) × c^{3}b^{5}a^{2}.
 #21: The two groups will have exactly the same
number of subsets! The 30th row of Pascal's
Triangle shows us the number of subsets from a
30element set: C(30,0), C(30,1), C(30,2), . . . ,
C(30,29), C(30,30). There are 31 such entries in
row 30, and the sum C(30,0) + C(30,2) + ... +
C(30,28) + C(30,30) is the number of subsets with
an even number of pliers. The sum C(30,1) +
C(30,3) + ... + C(30,27) + C(30,29) is the number
of subsets with an odd number of pliers. These two
sums are identical, each equal to (1/2) × 2^{30}
= 2^{29}. This is true for every row in
Pascal's Triangle. Can you justify why?
 #25: We can begin with the middle digit.
There are 10 options for this digit. Next, choose
the digit for the leftmost (millions) position.
This cannot be 0, so there are 9 options here.
Between the middle digit and the leftmost
(millions) digit ther are two places for digits.
Each can be filled in 10 ways. Once we have
selected digits for these four positions in the
number, we have no remaining choices if a
palindrome is required. Thus, by the
multiplication principle, there are 9 × 10^{3}=
9000 7digit palindromes.

