Math For Liberal Arts I

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:

1. 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?
2. 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 26·26 = 676 (since we have 26 possibilities for each letter). Similarly, the number of ways of choosing three letters is 26·26·26 = 17,576. Now we could choose two letters, or we could choose three letters, but we cannot choose both. This means that there are 676 + 17,576 = 18,252 ways to choose the letters for our license plate.

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 10·10·10 = 1000, and the number of ways of choosing four digits is 10·10·10·10 = 10,000, so that the number of ways to choose the digits for our license plate is 1000 + 10,000 = 11,000. Now the Fundamental Counting Principle tells us that the number of possible license plates is the number of possible ways of choosing the letters, times the number of possible ways of choosing the digits, namely 18,252·11,000 = 200,772,000.

(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 26·26·10·10·10 = 676,000. The number of license plates consisting of two letters followed by four digits equals 26·26·10·10·10·10 = 6,760,000. The number of license plates consisting of three letters followed by three digits equals 26·26·26·10·10·10 = 17,576,000. And the number of license plates consisting of three letters followed by four digits equals 26·26·26·10·10·10·10 = 175,760,000. Then the total number of possible license plates equals 676,000 + 6,760,000 + 17,576,000 + 175,760,000 = 200,772,000.)

3. 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)?
4. 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 5·4·3·2·1 = 120 (which, in the notation of the next section, we recognize as 5! = 5P5.)

#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 4·2·3 = 24.

#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 5·4·3·2·1 = 120. Only one of them is correct, so the number of ways of doing it incorrectly is 119.

#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 10·10·10 = 1000 possible triples of digits, only one of which is not allowed. Therefore, the number of possibilities for the first task is 1000 - 1 = 999. Second, we choose the two digits between the two dashes of a social security number. As in the first task, we are told that any two digits except 00 are allowed. Again there are ten choices for each of the two digits, so there are 10·10 = 100 possible pairs of digits, only one of which is not allowed. Therefore, the number of possibilities for the second task is 100 - 1 = 99. Finally, we choose the last four digits of a social security number. We are told that any four digits except 0000 are allowed. As before, there are ten choices for each of the four digits, so there are 10·10·10·10 = 10000 possible quadruples of digits, only one of which is not allowed. Therefore, the number of possibilities for the third task is 10000 - 1 = 9999. Putting it all together using the Fundamental Counting Principle, we see that the number of possible social security numbers is 999·99·9999 = 988,911,099.

#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 52·51·50·49·48 = 311,875,200. (In the notation of the next section, this is 52P5.)

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:  8! / 6! =(8·7·6·5·4·3·2·1) / (6·5·4·3·2·1) =8·7 = 56.

(d) SOLUTION:  Again no parentheses, but this time we do have to evaluate the factorials before we may subtract:  5! - 3! =(5·4·3·2·1) - (3·2·1) = 120 - 6 = 114.

(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:

1. 7P5 + 6P4=
SOLUTION:  By definition,  7P5= 7·6·5·4·3 = 2520.  (That is, we compute the product of 5 consecutive whole numbers, with 7 the largest.) Similarly,   6P4= 6·5·4·3 = 360  (the product of 4 consecutive whole numbers, starting at 6 and decreasing). Therefore,   7P5+ 6P4= 2520 + 360 = 2880.

2. 7P5 / 6P4 =

SOLUTION:  We already computed in part (a) that 7P5 = 7·6·5·4·3 = 2520  and   6P4= 6·5·4·3 = 360 , so we now have   7P5 / 6P4 =2520 / 360 = 7. Alternatively, we could cancel before multiplying out:  7P5 / 6P4 =(7·6·5·4·3) / (6·5·4·3) = 7.

3. 7P5 - 6P4=

SOLUTION:  Using the computations from part (a), we get   7P5 -6P4 = 2520 - 360 = 2160.

4. 7P5 · 6P4=

SOLUTION:  Again using the computations from part (a), we get   7P5 · 6P4= 2520·360 = 907,200.

#4, page 17.

1. 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 have   100! / 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: __________
Assuming that each player can play each position, in how many ways can he do this?

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  25P9.  (If we are interested in a numerical answer, we can easily compute that 25P9 =25·24·23·22·21·20·19·18·17 = 741,354,768,000.)

#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.
6P6 = 6·5·4·3·2·1 = 720
6P5 = 6·5·4·3·2 = 720
6P4 = 6·5·4·3 = 360
6P3 = 6·5·4 = 120
6P2 = 6·5 = 30
6P1 = 6
6P0 = 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  9P4= 9·8·7·6 = 3024.

#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.)