**Solutions to Homework §1.3, pages 12**-**13, **
**#1, 2, 6, 9, 10, 12**

**#1, page 12.** Work out the answers to Questions 5
and 6 at the beginning of Section 2:

- In a certain state, automobile license plates consist of either two or three letters, followed by three or four digits. How many different license plates are possible in that state?
- In the American League Central Division baseball standings, how many outcomes are possible without ties, so that one team is in first place, one in second place, one in third place, one in fourth place, and one in fifth place (there being five teams in all)?

**SOLUTION:** We use the Fundamental Counting Principle: We
choose some letters, and, having made this choice, we choose some digits.
For the first task, we could choose two letters, or we could choose three
letters. Because letters are allowed to repeat, the number of ways of choosing
two letters is

For the second task, we could choose three digits or we could choose
four digits. Because digits are also allowed to repeat, the number of ways
of choosing three digits is

(Alternatively, we could distinguish the four types of license plates,
count up the number of possibilities for each, and then add. The number
of license plates consisting of two letters followed by three digits equals

**SOLUTION:** There are 5 possibilities for which team finishes
first. Having chosen a first-place team, there are 4 possibilities for
which team finishes second. Having chosen these two, there are 3 possibilities
for which team finishes third. Having chosen these three, there are 2 possibilities
for which team finishes fourth. And finally, having chosen the first four
teams, there is only 1 possibility for which team finishes fifth. Using
the Fundamental Counting Principle, the number of possible outcomes is
_{5}P_{5}

**#2, page 12.** Suppose the Kruse family wants to make the
trip discussed in this section and then, as they return to Chicago, visit
one of the other three families not visited on the way down. How many round-trip
routes are there?

**SOLUTION:** We simply add a third task to the two already
discussed in this section. That is, the first task is to decide which of
the 4 families (Dallas, Hattiesburg, Louisville, or Philadelphia) they
will visit first. The second task is to decide which of the 2 cities (New
Orleans or Charleston) they will visit for vacation. Then the third task
is to decide which of the 3 remaining families (not chosen in the first
task) they will visit on the way home. Using the Fundamental Counting Principle,
the number of possible outcomes is

**#6, page 12.**

**SOLUTION:** Since there is clearly only one way to do this
*correctly*, the idea is to count the total number of ways he could
stuff the envelopes, and then subtract one. There are 5 possibilities for
which letter goes in the first envelope. After choosing a letter for the
first envelope, there are only 4 possibilities for which letter goes into
the second envelope. After choosing these two letters, there are still
3 possibilities for which letter goes in the third envelope. After choosing
these three, there are still 2 possibilities for which letter goes into
the fourth envelope. And finally, after choosing four letters, there is
only 1 possibility for which letter goes into the last envelope. Using
the Fundamental Counting Principle, the number of possible ways of stuffing
the envelopes is

**#8, page 12.**

**SOLUTION:** Let us distinguish the following three tasks,
in choosing a social security number. First, we choose the first three
digits of a social security number. We are told that any three digits *except*
000 are allowed. Now there are ten choices for each of the three digits,
so there are *except* 00 are allowed. Again there
are ten choices for each of the two digits, so there are *except* 0000 are allowed. As before,
there are ten choices for each of the four digits, so there are

**#9, page 13. **

**SOLUTIONS: **There are 4 • 3 • 2 • 1 = 24 distinct arrangements of the letters hi MORN. There are only 12 distinct arrangements of the letters in the word MOON, which can be found by listing:

MOON MONO MNOO OMON OMNO NMOO OOMN ONMO NOMO OONM ONOM NOOM

One can apply the counting principle in the following way. Let Task A consist of arranging the letters in MOON, and Task B changing one of the O's to an R. Task A followed by Task B results
in an arrangement of the letters of the word MORN. This can be done in 24 ways, so the number of ways to do Task A times the number of ways to do Task B must be 24. Since the number of ways to do Task B is 2, we see that the number of ways to do Task A must be 12. Try this line of reasoning on TEEPEE and QUEUE.

**#11, page 13.**

**SOLUTION:** Sam has 52 possibilities for the first choice.
Having made that choice, there remain 51 cards in the deck for the second
choice. Having made both of these choices, there are 50 cards left in the
deck for the third choice, and so on. Thus, using the Fundamental Counting
Principle, the number of ways of choosing five cards from a deck of 52,
and arranging them in a row, is _{52}P_{5}.)

**Solutions to Homework §1.4, pages 17-18, #1,
2, 4 (a), 5, 6, 9, 11 and §1.5, pages 22-23,
1(a)(b), 2, 5**

**#1, page 17.**

**(a) SOLUTION:** By definition: 5! =
5·4·3·2·1 = 120.

**(b) SOLUTION:** Careful, we have to evaluate *inside* the
parentheses first: (5 - 3)! =
2! = 2·1 = 2.

**(c) SOLUTION:** In the absence of parentheses, we compute the
factorials before dividing; on the other hand, we may cancel the common
factors in numerator and denominator before multiplying them out, which makes
the computation much easier:

**(d) SOLUTION:** Again no parentheses, but this time we do have
to evaluate the factorials before we may subtract:

**(e) SOLUTION:** Parentheses, so we evaluate them first: 4!
/ (4 - 3)! = 4! / 1! =
(4·3·2·1) / 1 = 24.

**(f) SOLUTION:** Piece of cake: 5 -
3! = 5 - (3·2·1) =
5 - 6 =-.

**#2, page 17.** Compute the following:

_{7}*P*_{5}+_{6}*P*_{4}=

**SOLUTION:**By definition,_{7}*P*_{5}= 7·6·5·4·3 = 2520. (That is, we compute the product of 5 consecutive whole numbers, with 7 the largest.) Similarly,_{6}*P*_{4}=6·5·4·3 = 360 (the product of 4 consecutive whole numbers, starting at 6 and decreasing). Therefore,_{7}*P*_{5}+_{6}*P*_{4}=2520 + 360 = 2880 .

_{7}*P*_{5}/_{6}*P*_{4}=**SOLUTION:**We already computed in part (a) that_{7}*P*_{5}= 7·6·5·4·3 = 2520 and_{6}*P*_{4}= 6·5·4·3 = 360 , so we now have_{7}*P*_{5}/_{6}*P* . Alternatively, we could cancel before multiplying out:_{4}=2520 / 360 = 7_{7}*P*_{5}/_{6}*P* ._{4}=(7·6·5·4·3) / (6·5·4·3) = 7

_{7}*P*_{5}-_{6}*P*_{4}=**SOLUTION:**Using the computations from part (a), we get_{7}*P*_{5}-_{6}*P*_{4}= 2520 - 360 = 2160.

_{7}*P*_{5}·_{6}*P*_{4}=**SOLUTION:**Again using the computations from part (a), we get_{7}*P*_{5}·_{6}*P*_{4}= 2520·360 = 907,200.

**#4, page 17.**

- What is 100! / 99! ? 101! / 100! ?
100! / 100 ?
**SOLUTION:**By definition, 100! / 99! = (100·99·98·...·2·1) / (99·98·...·2·1) (where the three dots "..." indicate some factors not shown). Now we see that each of the factors in the denominator cancels with the corresponding factor in the numerator, with only the factor of 100 left over in the numerator. That is, we have100! / 99! =(100·99·98·...·2·1) / (99·98·...·2·1) = 100 . Similarly,101! / 100! = (101·100·99·...·2·1) / (100·99·...·2·1) = 101 (after canceling the common factors of 100, 99, ..., 2, and 1). Finally,100! / 100 =(100·99·98·...·2·1) / 100 = 99·98·...·2·1, since now we have a factorial in the numerator only, so we have only the common factor of 100 to cancel. But we note that the product 99·98·...·2·1 is the product of the numbers from 99 down to 1, which we defined to be 99!. Therefore,100! / 100 =(100·99·98·...·2·1) / 100 =99·98·...·2·1 = 99! .

**#5, page 17.** A baseball team consists of 25 players. Barry
Griffey needs to fill in the 9 blanks in his starting lineup with names of
players from his team:

Pitcher: | __________ |

Catcher: | __________ |

1st Base: | __________ |

2nd Base: | __________ |

3rd Base: | __________ |

Shortstop: | __________ |

Left Field: | __________ |

Center Field: | __________ |

Right Field: | __________ |

**SOLUTION:** Since we assume that any player can play any position,
we see that there are 25 possibilities for who shall pitch. Having made that
choice, there are 24 possibilities left for who shall catch. Having made both of
these choices, there are still 23 possibilities left for who shall play 1st
base, and so on. Thus, the number of ways of assigning players to positions is
25·24·23·22·21·20·19·18·17, which we recognize as _{25}*P*_{9}.
(If we are interested in a numerical answer, we can easily compute that _{25}*P*_{9}
=25·24·23·22·21·20·19·18·17 =

**#6, page 18.** Work out all the numbers of permutations of
six things taken six at a time, five at a time, four at a time, etc., just as we
already did for seven things.

**SOLUTION:** Using the definition, we easily compute the following.

_{
6}*P*_{6} = 6·5·4·3·2·1 =
720

_{
6}*P*_{5} = 6·5·4·3·2 =
720

_{
6}*P*_{4} = 6·5·4·3 =
360

_{
6}*P*_{3} = 6·5·4 =
120

_{
6}*P*_{2} = 6·5 =
30

_{
6}*P*_{1} = 6

_{
6}*P*_{0} = 1 (by
convention!)

**#9, page 18.** Using only the nonzero digits, how many
four-digit numbers with distinct digits can be constructed?

**SOLUTION:** There are 9 nonzero digits, namely 1, 2, 3, 4, 5, 6,
7, 8, and 9. Any of these can be chosen for any spot in our four-digit number,
but no digit can be reused. Therefore, there are 9 possibilities for the first
digit, then there are 8 possibilities left for the second digit, then there are
7 possibilities left for the third digit, and finally there are 6 possibilities
left for the fourth digit. Thus, the number of four-digit numbers with distinct
nonzero digits is _{9}*P*_{4}=

**#11.** There are 5 • 4 • 3 = 60 numbers, of which 3 • 4 • 3 = 36
are odd. (Choose an odd digit for the position on the far right first.)