Math 3113, Section 4
Fall 1999
Solutions to Chapter 2 review questions
NOTE: It is also strongly recommended to review past homework problems
and your notes from group work in class. It is not claimed that
the problems on this page represent every type of question that could occur
on the test.

Area codes for telephone numbers obey the following rules: (I) the first
number cannot be 0 or 1; (ii) the second and third numbers can be any number
from 0 through 9, but (iii) the second and third numbers cannot be the
same.
Determine the number of possible area codes.
Solution: 8·10·9 = 720

With the same rules as in the previous problem:

Determine how many area codes have the first digit equal to 6.
Solution: 1·10·9 = 90

Determine how many area codes have the last digit equal to 6.
Solution: 8·10·1 = 80

Determine how many area codes have the first digit and the last
digit equal to 6.
Solution: 1·10·1 = 10

Determine how many area codes have the first digit or the last digit
equal to 6.

Suggestion: You may be able to use the formula n(AB)
= n(A) + n(B)  n(AB).
Solution: If we use the suggestion,
then it becomes 90 + 80  10 = 160.
Thus, we take A = {area codes with
first digit = 6} and B = {area codes with last digit = 6}.
So, AB
= {area codes with first and last digits = 6},
while AB
= {area codes with first digit = 6 or last digit = 6}

How many ways can 5 children be chosen from a group
of 12 and arranged in a row?
Solution: We can view it as
(12)_{5} = 12·11·10·9·8 or as
In either case, the numerical value is 95040.

How many ways can 5 children be arranged in a circle?
(Two arrangements are different if some child has a different person the
left or a different person on the right.)
Solution: We can choose one child's
place in the circle, and then arrange the other children in the remaining
4 places.
There are 4! = 24 ways to do this.

How many ways can 5 children be chosen from a group
of 12 and arranged in a circle.

Suggestion: Think of this as a process
with two steps. The two steps may involve methods from different sections
of the notes!
Solution: There are
ways to choose 5 children from a group of 12.
For each group, there are 4! = 24 ways to arrange them in a circle.
So, there are 792·24 = 19008 ways to accomplish
the assigned task.

Suppose that Bill lives k blocks south and nk
blocks west from work.
Explain why the number of possible routes Bill can
take to work is .
Solution: Bill has to walk nblocks
to work: kblocks north and nkblocks east. Choosing a route
is the same as making a "word" of nletters, of which kare
N's and nkare E's. So, we have to choose kof the npositions
in which to put the N's. The number of ways to do this is.

= Exercise 2.3.11 in the notes.
Method 1: A set with nelements
has a total of 2^{n} subsets. For each value of kbetween
0 and n(inclusive) there are
subsets with k elements. This gives the equation:
2^{n} =
+ +
+ ... + +
+ .
Method 2: Apply the binomial theorem
with xand yboth equal to 1, to obtain the same equation.

Use the binomial theorem to evaluate the following:

(x  y)^{5}
Solution: x^{5}  5x^{4}y
+x^{3}y^{2}
 x^{2}y^{3}
+ 5xy^{4}  y^{5},
or
x^{5}  5x^{4}y + 10x^{3}y^{2}
 10x^{2}y^{3} + 5xy^{4}  y^{5}
.

(3x + y)^{4}
Solution: 3^{4}x^{4}
+ 4·3^{3}x^{3}y +
·3^{2}x^{2}y^{2} + 4·3xy^{3}
+ y^{4},
or
81x^{4} + 108x^{3}y + 54x^{2}y^{2}
+ 12xy^{3} + y^{4}

= Exercise 2.3.39 in the notes.
Solution: I think that choosing a
team is just supposed to mean that we choose the members of the team, rather
than assigning them to particular positions.
So, this can be done in
ways.

= Exercise 2.3.40 in the notes.
Solution: I had thought that this
was the same as the previous problem, but one member of the class pointed
out another interpretation. Namely, we can completely reverse the two teams
in a given division, and still have the same division into two teams.
So, in this way the answer becomes .

How many rearrangements of the letters of the word MINNEHAHA
are there?
Solution: There are 9 letters altogether.
Proceed in stages, as follows:

There
ways to choose 2 places for the N's.

After choosing places for the N's, there are
ways to choose 2 places for the H's.

After choosing places for the N's and the H's, there
are ways to
choose 2 places for the A's.

Finally, there will be 3! = 6 ways to choose places
for the remaining 3 letters.

Altogether the number of rearrangements is ··
·3! = 36·21·10·6 = 45360.

...

In how many ways can 12 crayons of distinct colors be
distributed equally to 3 children?
Solution:··=
495·70·1 = 34650.

In how many ways can 12 crayons be divided into 3 sets
of 4 each? ?
Solution:···
= 34650/6 = 5775

= Exercise 2.4.16 in the notes.
Solution: Distributing the 10 quarters
to 5 people amounts to solving the equation
x_{1} + x_{2}
+ x_{3} + x_{4} + x_{5} = 10,
where the solutions are to be integers >
0. By exercise 2.4.14, the number of solutions is thus:
=
= =1001

= Exercise 2.4.17 in the notes
Solution: The given equation is x
+ y + z=13, with x> 1, y> 0, and z>
6. We "invent" new unknowns,
setting a = x 1, b = y,and
c =
z 6. Then we have to solve a + b + c =6, with a, b,and
call
>
0.
So, we can apply the result of exercise 2.4.14 again,
this time with n= 3 and k= 6.
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For questions: roberts@math.umn.edu