
Chapter 2 Review
 #3: (a) To have a pair, Tim needs a
lefthand glove and a righthand glove. There are
12 of each in the drawer. The worst case scenario
is that Tim first picks all 12 gloves that all are
for one hand. When he chooses a 13th glove,
however, he's assured of a pair. In fact, the
response is the same for (b), because, if Tim has
picked all 12 gloves for the same hand, he'll have
every color. The 13th pick assures that there will
be a color match as well as a left/right pair.
Chapter 3 Section 3
 #30:
 Situation I: Assume that order or
arrangement is significant. For example, a white
hat on Tim and a green hat on Tom is different
from a green hat on Tim and a white hat on Tom.
 (a) Each brother has 7 choices, so the
number of 5hat sets is 7×7×7×7×7 = 7^{5}.
 (b) P(7,5): The first brother has 7 colors
to choose from, the next has 6, and so on,
down to the last brother who has 3 colors to
choose from.
 (c) There are 7 colors for the hats.
Therefore, there are 7 ways the set of four
brothers could have chosen the same color hat.
There remain 6 colors for the fifth brother to
choose. Finally, there are 5 ways to choose
the one brother with a hat a different color
from the other. Therefore, this could have
occurred in 7×6×5 = 210 ways.
 Situation II: Assume that order or
arrangement is not significant. For example, a
white hat on Tim and a green hat on Tom is the
same as a green hat on Tim and a white hat on
Tom, because, from a hat perspective only, we
have a white hat and a green hat.
 (a) We need to determine the possibilities
for hat color combinations, and with each of
those determine the number of ways it could
occur. This includes: (i) All hats are the
same color. There are 7 colors, so there are 7
ways to have all 5 hats the same color. (ii)
Four (4) hats the same color and 1 hat that is
a different color. There are 7 ways to choose
a color for the four likecolored hats and 6
ways to choose a color for the fifth hat, so
42 in all. (iii) Three (3) hats the same color
and the other two hats match in a different
color. There are 7 ways to choose a color for
the three likecolored hats and 6 ways to
choose a color for the two hats that match a
different color, so 42 in all. (iv) Three (3)
hats the same color and the other two hats do
not match each other. There are 7 ways to
choose a color for the three likecolored
hats, and C(6,2) for the other two, so 7×15 =
105 in all. (v) Two (2) hats match color, two
other hats match color, and the fifth hat
differs from the first two colors. There are
C(7,2)×5 = 105 in all for this option. (vi)
Two hats match in color and the other three
hats are different from that and different
from each other. There are 7×C(6,3) = 140 in
all. (vii) All five hats differ in color.
There are C(7,5) = 21 ways for this to occur.
This yields a total of 462 options.
 (b) C(7,5)
 (c) There are 7 colors for the hats.
Therefore, there are 7 ways the set of four
brothers could have chosen the same color hat.
There remain 6 colors for the fifth brother to
choose. Therefore, this could have occurred in
7×6 = 42 ways.
Chapter 3 Section 4
 #1: C(20,3) = 20!/(17!3!) = 1140
 #3: C(12,6) = C(11,5) + C(11,6)
 #4: We know C(21,8) = C(20,7) + C(20,8), so
C(21,8)  C(20,7) = C(20,8).
 #7: Using Pascal's Formula, C(n,k) +
C(n,k+1) = C(n+1,k+1).
 #8: (r + s)^{3} = C(3,0)r^{3} +
C(3,1)r^{2}s + C(3,2)rs^{2} +
C(3,3)s^{3} = r^{3} + 3r^{2}s
+ 3rs^{2} + s^{3}
 #9: Because we have n repetitions of the
factor (x + y), the collectedterms
expansion will have terms that include from n
repetitions of x to 0 repetitions of x.
This means there are n+1 collected terms.
 #11: (a) 2^{12} uncollected terms (b)
When a occurs 5 times, we know (2a)^{5} is
a factor, as is (b)^{7}. We also know
there are C(12,5) = C(12,7) ways to have this pair
of factors occur. Therefore, the collected term is
C(12,5)*(2a)^{5}(b)^{7}= 792*32a^{5}*(1)b^{7}
= 25344a^{5}b^{7}. This shows
that H = 25,344.
 #13: (a) There are 4^{12}
uncollected terms. (b) The value of J is
(12!)/(5!1!0!6!) = 5544.
 #15: There are 5 + 1 = 6 collected terms.
 #17: (a) 4^{10} uncollected terms (b)
The value of K is (10!)/(3!1!2!4!) = 12,600.
 #19: C(7,0)*(4m)^{7}(3t)^{0} +
C(7,1)*(4m)^{6}(3t)^{1} +
C(7,2)*(4m)^{5}(3t)^{2}= 1*4^{7}*m^{7}*3^{0}*t^{0}
+ 7*4^{6}*m^{6}*3^{1}*t^{1}
+ 21*4^{5}*m^{5}*3^{2}*t^{2}
= 16,384m^{7} + 86,016m^{6}t
+ 193,536m^{5}t^{2}
 #23: (a) K = C(10,7) = C(10,3) = 120 (b) C(10,0)
+ C(10,1) + ... + C(10,9) + C(10,10) = 2^{10}
 #25: No. The exponents in the term do not sum to
12, and every term in that expansion must have
exponents that sum to 12.
 #29: (a) 5^{11} (b) K =
11!/(4!2!1!4!) = 34,650 (c) The only way to get 11
as a coefficient is for the multinomial
coefficient 11!/(e_{1}!e_{2}!e_{3}!e_{4}!e_{5}!)
= 11, where e_{1}+ e_{2}+ e_{3}+
e_{4}+ e_{5} = 11 and each e_{i}
is a nonnegative integer no greater than 11. This
can only occur when the denominator product is
10!, because 11!/(k) = 11 implies k = 10!. The
only way for the denominator to be 10!, with the
restrictions on e_{i} already
stated, is for one of the e_{i} to
be 10, another to be 1, and the other three to be
0. This can be done in 20 ways, because there are
5 ways to select the first e_{i}, 4
ways to select the next, and the remaining three
must then be used.
 #33: Solutions will vary.
 #35: (a) Here, for each of Seth's card
selections, he has 10 choices. Therefore, there
are 10^{5} possible ways for Seth to make
his selection. (b) Now Seth can choose no more
than one of a kind, and the order of selection is
not a factor. Therefore, there are C(10,5)
selections possible.
 #37: We know S(k) = 2^{k}. This gives us
S(12)  S(10) = 2^{12}  2^{10} =
2^{10}(2^{2}  2^{0}) = 2^{10}(4
 1) = 2^{10}(3) = 3*S(10).
 #38: Solution:
2[C(9,9)+C(10,9)+C(11,9)+…+C(18,9)] Think about
this as a modification or application of a
grid problem. A "path" passes through
integer ordered pairs (x,y), starting at the
origin (0,0) and ending at either (a) an
ordered pair on the line y = 10, with x an
integer ranging from 0 to 9, or (b) an
ordered pair on the line x = 10, with y an
integer ranging from 0 to 9. We just need to
count the paths, using the gridproblem
strategy, required to get to each of these
ending points. The only way
to get to (x,10) or (10,y), for x <= 9
and y <=9, is to get there from (x,9) or
from (9,y), respectively. That is, once the
value "10" is reached, there is no more
moving along the horizontal line y=10 or the
vertical line x=10. As soon as a team scores
10, the match is over. The picture below may
help illustrate this.

