1.

Respond to each of these questions. While you may show
steps leading to your solution, you do not need to generate
written explanations for questions (a) through (e) on this
page.
a) Express P(16,5) using factorial notation.
b) How many distinct arrangements exist for the letters
in the word reverberator?
c) In the expansion of ,
state:
i) the number of uncollected terms:
ii) the coefficient J in the collected term :
d) Determine the number of collected terms in the
expansion of .
e) Replace w, x, y, and z in C(12,6) = C(w,x) + C(y,z) to
illustrate Pascal's Formula, a fundamental relationship that
exists in Pascal's Triangle.
C(12,6) = C(11,6) + C(11,5)



2.

Thum lives in Grid City, where the streets are laid our
in a grid, running east/west and north/south. Thum's house
is in the northwest corner of the city and his girlfriend,
Bolina, lives in the southeast corner of the city. Thum's
house is 8 blocks north and 12 blocks west of Bolina's. The
image below shows the entire city.
a) How many 20block paths are there from Thum's to
Bolina's, assuming all streets exist and are open to
traffic?
Traveling from Thum's to Bolina's in a 20block
path requires traversal of 8 southbound streets and
12 eastbound streets. Thus, we need to look at all
permutations of the "word" SSSSSSSSEEEEEEEEEEEE to
determine the number of paths.
The number of unique arrangements of this
20character word is
C(20,8)=C(20,12)=20!/(12!8!).

b) Grid City eventually will build a walking mall in the
shaded location shown below, thereby eliminating a oneblock
length of street. Under these conditions, how many 20block
paths are there from Thum's to Bolina's?
From the C(20,8) paths determined in (a), we
must subtract all paths that contain the restricted
street, shown above as going from x to y.
Using the same reasoning as in (a), there are
C(9,4) paths from T to x, 1 path from x to y, and
C(10,4) paths from y to B. Because each of these
legs (T to x, x to y, and y to B) are independent
from each other and each must be included in a
20block path from T to B through the restricted
street, we multiply the three values to determine
the total number of 20block paths from T to B
through the restricted street. This is
C(9,4)*1*C(10,4)=C(9,4)*C(10,4). We subtract this
from C(20,8) to determine the number of paths we
can use to go from T to B and avoid the restricted
street.
The resulting number of paths is
C(20,8)C(9,4)*C(10,4).



3.

Referring to the letters in the word ACMAESTHESIA,
solve each of the following problems. Each problem is
independent and separate from the others.
a) How many unique arrangements are there for the letters
in this word?
There are 12! permutations of the letters in the
word, but we must account for duplication and
therefore divide by 3!2!2! because there are 3
identical As, 2 identical Es, and 2 identical Ss.
There are 12!/(3!2!2!) unique arrangements for the
letters in this word.

b) How many unique arrangements exist if two or more
vowels cannot be adjacent to one another?
Begin by placing the 6 consonants in the word.
Accounting for two Ss, this can be done in 6!/2!
ways. This leaves us 7 places among the 6
consonants in which to place vowels so no vowels
will be adjacent. We first choose 6 of the 7 places
for the vowels, done in C(7,6) ways. We now permute
the vowels within the places chosen. this can be
done in 6!/(3!2!) ways, accounting for duplicate As
and Es. We multiply the results to determine the
total number of ways of arranging the letters in
the word so no vowels are adjacent. This is
(6!/2!)*C(7,6)*[6!/(3!2!)].

c) If the only distinction we can make is between
consonants and vowels, how many arrangements can be
made?
With no distinction among the vowels or among
the consonants, we have 12 total objects of just
two types, 6 of type "consonant" and 6 of type
"vowel." We permute the 12 objects and account for
the duplicates among the types. This results in
12!/(6!6!) ways to make this sort of
arrangement.

d) If the only distinction we can make is between
consonants and vowels, and two or more consonants cannot be
adjacent to one another, how many arrangements can be
made?
Again with no distinction among the vowels or
among the consonants, we have 12 total objects of
just two types, 6 of type "consonant" and 6 of type
"vowel." We first place the vowels, and because
they are indistinguishable, there is 1 way to do
that. We create 7 spaces among the vowels, and we
choose 6 of these for the consonants. This is just
C(7,6), our desired result. There are 7
arrangements that can be made under these
conditions.



4.

Louise invests her money in $200 lots. She has $3000 to
invest and her daughter Gina has suggested five different
mutual funds for Louise's investments.
a) How many different ways can Louise invest her money if
she insists on putting at least $200 in each of the five
funds her daughter recommended and uses only these five
funds?
Note that Louise has 15 lots of $200 each to
invest. Thus we are dealing with the placement of
15 objects that are identical.
If Louise drops 1 lot into each of the five
funds, she has 10 lots remaining to distribute.
This distribution of 10 lots can be done in any way
among the five funds. This is analogous to solving
the equation A+B+C+D+E=10 where A,B,C,D, and E must
be nonnegative integers. This placement can be done
in C(10+51,51)=C(14,4) ways.
In terms of objects and dividers, we have 10
objects and 4 dividers to permute. This can be done
in 14!/(10!4!) ways.

b) If Louise restricts her investments to these five
funds but may choose to not invest any money in one or more
of the funds, how many different ways can Louise invest her
money?
Again we have 15 lots of $200 each to
invest.
This distribution of 15 lots can be done in any
way among the five funds. This is analogous to
solving the equation A+B+C+D+E=15 where A,B,C,D,
and E must be nonnegative integers. This placement
can be done in C(15+51,51)=C(19,4) ways.
In terms of objects and dividers, we have 15
objects and 4 dividers to permute. This can be done
in 19!/(15!4!) ways.



5.

Consider the binomial .
a) Determine the number of uncollected terms in the
expansion of this binomial.
For the expansion of a binomial, this is just
2^t.

b) Show and describe a typical collected term in the
expansion of this binomial.
A typical term in the expansion will be
,
where x+y=t and C=t!/(x!y!).

c) If k = x, m = 2y, and t=8, determine N in the
collected term Nx^3y^5.
We are expanding the binomial (x+2y)^8. for the
collected term in question, we will have
C(8,3)*(x)^3*(2y)^5, which simplifies to .
Therefore, .



6.

A bag contains a virtually unlimited supply of red
marbles, blue marbles, white marbles, and yellow marbles.
Marbles of any one color are indistinguishable from each
other.
a) Marbles are drawn from the bag without looking until a
set of 6 marbles is created. How many different 6marble
sets could be created?
We have an essentially unlimited supply of four
different objects, so as we draw from the bag we
could generate a set of 6 marbles with various
numbers of each color marble, as many as 6 of one
color and as few as 0 of one or more colors.
This is analogous to solving the equation
R+B+W+Y=6 where R,B,W, and Y must be nonnegative
integers. This distribution can occur in
C(6+41,41)=C(9,3) ways.
In terms of objects and dividers, we have 6
objects and 3 dividers to permute. This can be done
in 9!/(6!3!) ways.

b) Marbles are drawn from the bag without looking until a
set of 24 marbles is created. Of all the 24marble sets we
could create, how many have at least two marbles of each
color?
Again we have an essentially unlimited supply of
four different objects, so as we draw from the bag
we generate a set of 24 marbles with various
numbers of each color marble. We are restricted to
having at least two of each color marble.
If we preselect 2 marbles of each of the four
colors, we have 16 marbles remaining to select.
This selection of 16 marbles can be done in any way
among the four colors. This is analogous to solving
the equation R+B+W+Y=16 where R,B,W, and Y must be
nonnegative integers. This distribution can occur
in C(16+41,41)=C(19,3) ways.
In terms of objects and dividers, we have 16
objects and 3 dividers to permute. This can be done
in 19!/(16!3!) ways.

c) Marbles are drawn from the bag without looking until a
set of 30 marbles is created. Of all the 30marble sets we
could create, how many have marbles of exactly three colors
in them?
Again we have an essentially unlimited supply of
four different objects and we draw from the bag to
generate a set of 30 marbles with various numbers
of each color marble. We are restricted to
30marble sets that contain exactly three colors.
This is equivalent to saying in the 30marble set
exactly one color is NOT represented.
If we preselect red as the marble color not
selected, we have 30 marbles to select from among
three colors. This selection of 30 marbles can be
done in any way among the three colors, with the
restriction that each of the remaining three colors
is represented by at least one marble. This is
analogous to solving the equation B+W+Y=30 where
B,W, and Y must be positve integers. This
distribution can occur in C(301,31)=C(29,2)
ways.
In terms of objects and dividers, we have 30
objects and 2 dividers to work with. Because all
remaining colors among the three must be
represented, the 2 dividers must be placed
somewhere among the 29 spaces between the 30
objects. This can be done in C(29,2) ways.
The same logic and calculations will result for
each of the other three cases, that is, when blue
is the preselected absent color, when white is the
preselected absent color, and when yellow is the
preselected absent color. Therefore, of all the
30marble sets we could create, there are 4*C(29,2)
with exactly three colors in them.


