Math 3113, Section 4
Fall 1999
Solutions to selected exercises

Exercise 2.1.3(Double
letter license plates and double number license plates) This was
a class exercise, but it's being presented as background for exercise 2.1.4,
a homework exercise which was more confusing.

Double letters (including triples). According
to the exercise, a double letter is where two adjacentletters are
the same; but we also have to include the possibility of triple letters.
And, presumably, the same applies for double numbers. So, we have these
possibilities, where "S" stands for "same" and "D" stands for "different ":

SSD### 26·1·25·10·10·10
= 650,000possibilities

DSS### 26·25·1·10·10·10
= 650,000possibilities

SSS### 26·1·1·10·10·10
= 26,000 possibilities
So, altogether there are 650,000 + 650,000 + 26,000
= 1,326,000 possibilities for a license plate with a double letter.

Double numbers (again, including triples) Here,
the possibilities for the letters will always be 26·26·26
= 17,576. Combining these with the possibilities for the numbers, we get
the following:

First 2 numbers the same: 26·26·26·10·1·9
= 1,581,840 possibilities

2^{nd} and 3^{rd} numbers the same:
26·26·26·10·9·1 = 1,581,840 possibilities

Triple number: 26·26·26·10·1·1
= 175,760 possibilities
Altogether, there are 1,581,840 + 1,581,840 + 175,760
= 3,339,440 possibilities for a license plate with a double number.

Exercise 2.1.4(License
plates which have double letters and/or double numbers)

Double letter anda double number.
Since the question involves "and",we use the
multiplication principle. Using "S" for "same" and "D" for "different"
in the letters, and #s and #d for "same" and "different" among the numbers,
we have these possibilities:

SSD#s#s#d: 26·1·25·10·1·9
= 58,500 possibilities

SSD#s#d#d: 26·1·25·10·9·1
= 58,500 possibilities

DSS#s#s#d: 26·25·1·10·1·9
= 58,500 possibilities

DSS#d#s#s: 26·25·1·10·9·1
= 58,500 possibilities

SSS#s#s#d: 26·1·1·10·1·9
= 2,340 possibilities

SSS#s#s#d: 26·1·1·10·9·1
= 2,340 possibilities

SSD#s#s#s: 26·1·25·10·1·1
= 6,500 possibilities

DSS#s#s#s: 26·25·1·10·1·1
= 6,500 possibilities

SSS#s#s#s: 26·1·1·10·1·1
= 260 possibilities
Finally (gasp!), that seems to be everything.
Adding all of these numbers, we see that there are 251,940 possibilities
for a license plate with a double letter anda double number.

Double letter ora double number
Here, the question involves "or", so that we can use the
addition principle. The following is a kind of "shortcut". Namely, let's
use the formula: n(A B) = n(A) + n(B)  n(A B), from page 31 of
the notes. We take:

A = {all license plates with a double letter}, and

B = {all license plates with a double number}. Therefore:

A B = {all license plates with a double letter ora
double number}, and

A B = {all license plates with a double letter anda
double number}.
So, the expression n(A) + n(B), i.e.simple addition
would count the license plates with a double letter anda
double number twice. But the formula corrects for this by subtracting off
the double count. Thus:n(A B) = n(A) + n(B)  n(A B), or:
n(A B) = 1,326,000 + 3,339,440  251,940 = 4,413,500
possibilities for a license plate with a double letter ora
double number.

Exercise 2.3.31 [A
class exercise] (Choosing 4 students from a
group of 10, and then choosing a chairperson, vice chairperson, secretary,
and treasurer from these four.)

There are
ways to choose a selection of 4 students from a group of 10.

For each selection of 4, there are 4! ways to arrange
them in an ordered list.

Hence, there are ways
in which this process can be done.

Exercise 2.3.32 [A
class exercise] (Choosing a chairperson, vice
chairperson, secretary, and treasurer from a group of 10 students).

There are 10 choices for the chairperson. For each choice
of chairperson, there are 9 choices for the vice chairperson. (Etc. …)

So, the result is that there are (10)_{4} =
10·9·8·7 ways to do this process.

Exercise 2.3.33 [A
class exercise] (A formula
for . )

The first step is to observe that there are the same
number of possibilities for the process described in exercise 31 as for
the process described in exercise 32. Thus, there are exactly as many ways
to name a slate of 4 officers and then assign roles as there are ways to
elect the 4 officers in a stepbystep process.

So, we set the two answers equal, getting the following
equation:
= 10·9·8·7.

We solve this equation, thus finding the formula:
.

Exercise 2.3.34 [A
homework problem] (Formulas
for . )

As suggested in the notes, we use exercise 33 (above) as a model.
So, (n)_{k }, the number of permutations of nthings taken
kat
a time can be given as (i) one of the formulas from §2.2 and
(ii)
also
as .
(In
the second viewpoint, there are
ways to choose a subset of kelements, and then k!ways to
arrange its elements.)

From exercise 2.2.6, we have the formula (n)_{k}
= n·(n1)·(n2)···(n  k+1).
We again set the two answers equal, obtaining the following equation:
= n·(n1)·(n2)···(n  k+1).

We solve this equation, thus finding the formula:
.
(Version 1)

To derive the other version of the formula, we can multiply
numerator and denominator of the last fraction by (nk)!. Thus:
.
{Probably, you could say that the factor (nk)!
was chosen by "working back from the answer", but after doing that to guess
the answer, we can see that the algebra is valid.} We handle the numerator
in the new fraction by noting that:
[n·(n1)·(n2)···(nk+1)]·(nk)!
= [n·(n1)···(nk+1)]·(nk)·(nk1)···2·1,
which is just n ! So, we can rewrite this
formula as:
. (Version 2)
For questions
roberts@math.umn.edu
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