1.

Set N contains the following digits:
{0,1,2,3,4,5,6,7,8,9}; set L contains the following letters:
{A,B,C,D,E,F}; set S contains the following symbols:
{<,>,=}
(a) A unique 2character code is to be created by
selecting one digit from set N and one letter from set L. If
it matters not whether a digit or a letter is listed first,
how many unique 2character codes can be created? (3
points)
Solution: 2*10*6=120 codes
There are 10 digits to choose from followed by
six letters. Double that to change the order so
that a letter appears first.

(b) Mathilda drew on paper a lineup of six symbols from
S. Her lineup included 2 of < and 4 of >. How many
different 6symbol lineups containing 2 of < and 4 of
> could she create? (3 points)
Solution:C(6,2)=C(6,4)=15 ways
This is analogous to the S/D problem, asking how
many birth orders are possible in a family with 4
sons and 2 daughters. Mathilda created a 6symbol
string that included 2 of one type of symbol and 4
of another.

(c) Jackie was asked to create a subset of set N. She
could choose no more than one of each digit, her subset was
required to contain at least one digit, and her subset could
contain no more than 9 digits. How many different subsets
could Jackie create? (4 points)
Solution: 2^10  2 = 1022 possible subsets
There are 2^10 possible subsets of N, for we
have two choices with each element of N: include it
in a subset or not. The restrictions of the problem
means we cannot create the empty set nor can we
take the entire set N. Therefore, we must subtract
two subsets from the 2^10 that are possible.


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2.

2) The passenger door that is part of a keyless entry
system on a new car has a 5&endash;pad keypad on the
driver's door similar to the diagram shown here:
Every car is assigned a keyless entry code as it rolls
off the assembly line. If each code is a threedigit number,
such as 578, how many different keypad entries (unique
sequences of keypad pushes) are possible? For instance, the
code 578 has this keypad sequence:
Solution:5^3=125 unique keypad entries
Each code requires three keypad pushes. There
are 5 keypads available for each push, so there are
5*5*5 different ways to make a threestep keypad
entry.


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3.

Five pairs of shoes were in a single line in Winnie's
closet. Each pair was a solid color and there were five
different colors: Black, White, Brown, Red, and Green. Each
pair contained a left shoe and a right shoe, each
distinguishable from one another.
(a) With no regard to matching pairs of shoes, how many
ways exist for the shoes to be lined up in a single line? (3
points)
Solution: P(10,10)=10! ways
All the items are distinguishable from another,
so this is the permutations of 10 things taken 10
at a time.

(b) Suppose that four single shoes are chosen from these
five pairs. How many ways exist for these four single shoes
to contain only shoes for the left foot? (3 points)
Solution: C(5,4)*C(5,0)=5*1=5 ways
There are 5 leftfoot shoes and 5 rightfoot
shoes. We seek 4 of the five leftfoot shoes and
none of the five rightfoot shoes. We are choosing
not arranging, so we use combinations.

(c) Suppose that three single shoes are chosen from these
five pairs. Of all the ways to select three single shoes,
what portion of those will include a matched pair of shoes?
(4 points)
Solution: 40/120 = 1/3 of all threeshoe sets
will contain a matched pair
There are C(10,3)=120 ways to select 3 shoes
from the 10. To get one pair among the three that
are chosen, we have 5 pairs to choose from to get
the pair, matched up with one of the 8 remaining
shoes in the closet. This results in 5*8=40 3shoe
sets that contain a matched pair.


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4.

Consider the letters in the word
ACANTHOCHEILONEMIASIS.
Note: There are 21 letters in the word,
including 3 A, 2 C, 2 E, 2 H, 3 I, 1 L, 1 M, 2 N, 2
O, 2 S, and 1 T.

a) How many unique arrangements are there for the letters
in this word? (2 points)
Solution: 21!/(3!2!2!2!3!2!2!2!)
arrangements
There are 21! ways to arrange the 21 letters,
but there is duplication of letters. We divide by
3!2!2!2!3!2!2!2! to eliminate all duplicates.

b) How many arrangements exist if each arrangement must
begin and end with a consonant? (2 points)
Solution: (11*10*19!)/(3!2!2!2!3!2!2!2!)
arrangements
There are 11 consonants, so there are 11*10 ways
to arrange the first and last letters as
consonants. There are 19! ways to arrange the
reamining 19 letters. We divide by the same
denominator as in (a) to eliminate duplicates.

c) How many arrangements exist if all vowels must be kept
together? (3 points)
Solution: (10!12!)/(3!2!2!2!3!2!2!2!)
arrangements
Treat the 10 vowels as a chunk. There are 10!
ways to arrange the vowels within that chunk. With
the 11 consonants and the chunk of vowels, there
are 12 things to place which can be done in 12!
ways. We still divide by the same denominator as in
(a) and in (b) to account for duplication.

d) How many 5letter sets can be created using only the
unique letters in the word? (3 points)
Solution: C(11,5) sets
There are 11 unique letters in the word. We
select 5 of them. Arrangenment is not an issue
here, only selection, so we use a combination.


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5.

Three people who frequented a local juice bar were such
bitter enemies that they could not be trusted to sit on bar
stools that were next to each other. The juicebar
proprietor required that there always be at least one bar
stool (occupied or not) between any two of these bitter
enemies.
(a) What is the minimum number of bar stools, all in a
line, that is required to meet the proprietor's seating
restrictions for the three enemies? (2 points)
Solution: 5 stools
Each of the three enemies needs a stool and
there must be a stool between each pair: E1, S1,
E2, S2, E3. Because the enemies can occupy the
"outside" stools, we need two to place between
them. Thus we need 5 stools in all.

(b) How many ways could the three enemies be seated,
under these restrictions, if there were 8 bar stools in a
line? (4 points)
Solution: P(6,3)=(6,3)*P(3,3)=6*5*4=120 ways
We know that 5 of the 8 stools will not be
occupied by enemies, so we place these first. We
assume the stools are not distinguishable from one
another, so there is only one way to place these 5
stools. These five stools create 6 spaces among
them:
___ S1 ___ S2 ___ S3 ___ S4 ___ S5
___
We now arrange the three enemies among these 6
empty spaces, so there are
P(6,3)=(6,3)*P(3,3)=6*5*4 ways to arrange the
enemies.

(c) Generalize the solution to (b) for N enemies and B
bar stools. State any restrictions on the quantities N and
B. (4 points)
Solution: P(BN+1,N) ways
We place the BN stools that will not contain
enemies. This creates BN+1 empty spaces at which
the enemies can sit. Any choice of N of these BN+1
stools will meet the proprietor's requirements, so
there are P(BN+1,N) ways to seat the N
enemies.
Regarding restrictions, we must have
BN+1>=N, so B>=2N1. Try this with part
(a).


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6.

6) Oliveras approached her discretemathematics
instructor and showed him the following claims:
(a) C(n,r)=P(n,r)/r!
(b) P(n,r)=n!/(r1)!
State whether each claim is always true, sometimes true,
or never true. Show appropriate evidence to support your
response. In addition, if a claim is sometimes true, show a
case for which the claim is true and a case for which the
claim is false. If a claim is never true, include a case for
which the claim is false.(5 points each)
(a) C(n,r)=P(n,r)/r!
Solution: always true
Compare the leftside expression to the
rightside expression in this equation:
LS: C(n,r)=n!/(r!(nr)!)
RS:
P(n,r)/r!=(n!/(nr)!*(1/r!)=n!/(r!(nr)!)
We have shown that the right side can be made
equivalent to the left side, so the equation will
always be true.

(b) P(n,r)=n!/(r1)!
Solution: sometimes true
Compare the leftside expression to the
rightside expression in this equation:
LS: P(n,r)=n!/(nr)!
RS: n!/(n1)!
The RS will equal the LS only if (nr)!=(r1)!.
These two products (factorials) will only be true
if nr=r1, or if n=2r1. This is the condition on
n and r that muct hold for this result to be true.
Otherwise, it will be false.
Example: True Case: Let n=5 and r=3. Note that
2r1=n. P(5,3)=5*4*3, and 5!/2!=5*4*3. LS=RS=60
Example: False Case: Let n=3, r=3. Note that
2r1<>n. P(3,3)=3!, but 3!/2!=3. The RS is
not equal to the LS.


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