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251solnI 2/23/06

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I. Permutations and Combinations.

1. Counting Rule for Outcomes.

Text 4.38, 4.41, 4.46 [4.36*, 4.39*, 4.43*]

2. Permutations.

4.47 [4.44*], D&C pg. 97( pg. 85 in 3 rd

ed) 19.

3. Combinations.

4.48 [4.45*], D&C pg. 97 20-21, 29-34. I1-I4.

------------------------------------------------------------------------------------------------------------------------------------------------------------------

Exercise 4.38 [4.36 in 9 th

](Not in 8 th

edition):

If there are 10 multiple-choice questions on an exam, each having 3 possible answers, how many different possibilities are there for sequences of correct answers?

Solution:

3

10

59049

Exercise 4.41 [4.39 in 9 th

](Not in 8 th

edition):

A brand of women’s jeans can be ordered in seven different sizes, 3 different colors and 3 different styles. How many jeans have to be ordered if the store wants to have one pair of each type.

Solution:

7

3

3

63

Exercise 4.46 [4.43 in 9 th

] (Not in 8 th

edition):

A gardener has 6 rows in his garden available for 6 different vegetables. If each vegetable gets one row, how many different ways are there to position the vegetables in the garden?

Solution:

6 !

6

5

4

3

2

1

720

Exercise 4.47 [4.44 in 9 th

] (Not in 8 th

edition):

The Big Triple consists of picking the correct order of finish of the first 3 horses in the 9 th

race. If there are 12 horses entered in the race, how many outcomes are there?

Solution:

P r n

n

n

!

r

!

, so

P

3

12

12 !

9 !

12

11

10

1320

Downing and Clark, pg. 85, Application 19:

The number of ways four aces can be located in a deck of 52 cards..

Solution:

Apparently, order counts in this case.

P

4

52

52

52

!

4

52 !

  

!

52

51

50

49

6497400 .

Exercise 4.48 [4.45 in 9 th

] (Not in 8 th

edition):

The Quinella consists of picking the horses that will place first or second regardless of order. If 8 horses are entered in a race, how many winning combinations are there?

Solution:

C r n

n

n

!

r

!

r

!

, so

C

8

2

8

8 !

2

  

6

8

!

!

2 !

8

7

2

1

28

Downing and Clark, pg. 85, Application 20:

The sock drawer contains 4 red, 3 blue, and 2 brown socks, for a total of 9 socks. You pick 2 at random and want to know the probability of getting a pair of matching socks.

Solution:

Identify the following events:

R

2

red socks

,

Bl

2

blue socks

and

Br

2

brown socks

.

We want

P

R

Bl

Br

C

2

4

C

0

3

C

0

2

C

9

2

C

0

4

C

2

3

C

0

2

C

9

2

C

0

4

C

0

3

C

2

2

C

9

2

4

2

3

1

1

1

1

9

8

3

1

2

1

1

1

1

10

36

5

18

.

27778

251solnI 10/09/03

Downing and Clark, pg. 85, Application 21:

If we pick 2 coins from 5 pennies, and 4 dimes, what is the chance of getting 20 cents?

Solution:

The event only happens if we pick 2 dimes. There are

C

9

2

ways to pick 2 coins from 9, but only

C

4

2

ways to pick 2 dimes from 4 (and nothing from the 5 pennies).

C r n

n

n

!

r

!

r

!

. So

P

2

dimes

C

4

2

C

5

0

C

2

9

2

4

!

!

2 !

9 !

5 !

5 !

0 !

4

3

2

9

1

8

2

1

4

3

9

8

12

72

1

6

.

1667

2 !

7 !

Downing and Clark, pg. 85, Application 29:

If you have 20 blue balls and 30 red balls in a box, and you randomly pull out 20 balls, what is the probability that they will all be blue?

Solution:

There are 20 blue balls and we need all of them. There is only

C

20

20

1 way to do this. We divide this by the number of ways to take 20 balls from a total of 50.

C

50

20

50

50 !

20

!

20 !

50

49

48

47

20

19

46

45

18

17

44

16

15

43

42

14

13

41

12

40

39

11

10

38

9

37

8

7

36

35

6

5

34

33

4

3

2

1

32

31

50 !

!

20 !

1 .

7619

10

13

. So the answer is

1

C

50

20

1

1 .

7619

10

13

2 .

1

10

14

. There is another way to argue this. On the first try the chance of getting a blue ball is

20

50

. On the second try, the conditional probability of getting a blue ball, given that you got a blue ball on the first try is

19

49 because there are 19 blue balls left among the 49 remaining balls. This goes on until the 20 th

try, when there is only one blue ball left out of 31 remaining. If you multiply these 20 fractions together you get

50

49

20

48

19

47

18

46

17

16

45

44

15

14

43

13

12

42

41

11

40

10

39

9

38

8

37

7

6

36

5

35

4

3

34

2

1

33

32

31

1

C

50

20

.

Downing and Clark, pg. 85, Application 30:

Explain intuitively

C

0

n

.

Solution:

There is only one way to take nothing from

n

items (if order doesn't count).

C r n

n

n

!

r

!

r

!

, so

C

0

n

n

n

!

0

!

0 !

n

!

!

1

1 .

Downing and Clark, pg. 85, Application 31:

Explain intuitively

C

1

n

.

Solution:

There is

C

1

n

n

n

!

1

!

1 !

n

n n

ways to take 1 item from

n

n

n

1

1

!

!

1 !

. items, since there are

n

choices.

Downing and Clark, pg. 85, Application 32:

Explain intuitively

C n n

.

Solution:

There is only one way to take all from

n

items.

C r n

n

n

!

n

!

n

!

n

!

0 !

n

!

n

!

1

n

!

1 .

251solnI 10/09/03

Downing and Clark, pg. 85, Application 33:

Explain intuitively

Solution:

There are

n

ways to take and there are

n n

1 items from

ways to pick that one item.

C n n

1

n

n

n

!

1

  

n

C

n n

1

1

!

.

n

items. This is because we must leave out one item

1 !

n n

!

1

!

n

1

n n

1

1

!

!

n

.

Downing and Clark, pg. 85, Application 34:

Compare

C n j

and

C n n

j

. Explain intuitively.

Solution:

The number of ways to take take

n

j

items from

n j

items from

n

items must be the same as the number of ways to items since, in the first case we must choose

j

items to leave in, and . in the second case we must choose

j

items to leave out. Mathematically,

C n j

n

n

!

j

!

j

!

and

C n n

j

n

n

j n

!

  

n

j n

!

   

n

j

!

. These are the same except for the order of computation.

Problem I1:

What is the probability of two jacks in a hand of ten cards?

Solution:

The denominator must be the number of ways to take 10 cards from 52, numerator must be the number of ways to take 2 jacks from the four in the deck,

C

C

52

10

2

4

, while the

6 , multiplied by the number of ways to take the remaining 8 cards from the 48 non-jacks in the deck,

C

8

48

. Recall that

C r n

n

n

!

r

!

r

!

, so

P

2

jacks

C

4

2

C

52

C

10

8

48

2 !

4 !

2

42

!

52

!

48

40

!

10

!

!

!

8 !

4

3

2

52

1

51

48

50

47

46

45

44

43

42

8

49

7

48

6

5

47

4

46

3

2

45

1

44

41

43

10

9

8

7

6

5

4

3

2

1

. If we cancel the numbers 8 down through 1 and the numbers 48 down through 43 within both the numerator and the denominator, we get

P

2

jacks

52

6

42

51

50

41

49

6

42

41

52

10

51

9

50

49

6642

46410

.

1431 .

10

9

Problem I2:

What is the probability of five jacks in a hand of ten cards?

Solution:

This is impossible because there is no way to find 5 jacks in a single deck of cards. No amount of mathematics will help you here.

P

5

jacks

0 .

251solnI 10/09/03

Problem I3:

What is the probability of two jacks and three tens in a hand of ten cards?

Solution:

The denominator must be the number of ways to take 10 cards from 52,

C

52

10

, while the numerator must be the number of ways to take 2 jacks from the four in the deck,

C

2

4

6 , multiplied by the number of ways to take 3 tens from the four in the deck,

C

3

4

4

, and multiplied by the number of ways to take the remaining 5 cards from the remaining 44 cards that are neither jacks or tens,

C

44

5

. Recall that

C r n

n

n

!

r

!

r

!

, so

P

2

jacks

, 3

tens

C

2

4

C

3

4

C

5

44

52

C

10

4 !

4 !

44 !

2 !

2 !

3 !

1 !

39 !

5 !

52 !

52

4

3

51

2

1

50

4

49

44

43

42

41

48

5

4

47

3

46

2

1

45

40

44

10

9

8

7

6

5

4

3

2

1

43

. If we cancel the

42 !

10 !

numbers 5 down through 1 and the numbers 44 down through 43 within both the numerator and the denominator, we get

P

2

jacks

, 3

tens

52

4

3

51

2

1

50

4

49

42

48

41

47

40

46

10

9

8

7

6

45

52

42

41

51

5

49

8

2

7

6

47

46

564816

2809475760

.

00021

6

4

42

52

51

50

41

40

10

49

48

47

9

7

6

46

45

The last line involves a great deal of canceling, so I urge you to check my work and see what you get.

Problem I4:

.In a poker hand (5 cards), what is the probability of getting two hearts or two face cards (or both).

Solution: This is a very difficult problem. The only reason that it is here is to remind you of the considerations that have to be watched when you compute the probability of a union.

Let us define the following events:

H

2

Two hearts in hand

F

2 Two face cards in hand

Then, by the addition rule,

P

H

2

F

2

P H

2

H

2

F

2

.

Now, a deck of 52 cards contains 13 hearts and 12 face cards. Of these:

10 are nonheart face cards;

9 are nonface hearts

3 are hearts with faces.

Then

P

 

 

9880

2598960

C

30 cards that are neither face cards or hearts

13

2

C

3

39

C

5

52

 

9139

2598960

712842

2598960

.

274280 and

652080

2598960

.

250900

P

 

C

12

2

C

40

3

C

5

52

. The hard part is finding the intersection of these two events. This is because there are three different patterns by which it can occur. Let us list these with the number of ways that each pattern can occur.

251solnI 10/09/03

C

10

2

C

9

2

C

3

0

C

30

1

(a) Two nonface hearts, two nonheart face cards, no heart face cards. This can happen in

    

48600 ways.

.

C

10

1

C

9

1

C

3

1

C

30

2

(b) One nonface hearts, one nonheart face cards, one heart face cards. This can happen in

    

117450 ways.

C

10

0

C

0

9

C

2

3

C

3

30

1

(c) No nonface hearts, no nonheart face cards, two heart face cards. This can happen in

4060

12180 ways.

We can conclude that there are 48600+117450+12180=178230 ways to get both 2 hearts and two face cards. The probability is thus

P

F

2

H

2

178230

.

068577 .

Incidentally, we could not get this by

2598960 multiplying the probability of two face cards by the probability of two hearts because H2 and F2 are not independent as is demonstrated by

P

F

2

H

2

178230

712842

.

250027

.

assuming independence would be small. We thus conclude that

P

    

H

2

F

2

.

274280

.

250900

.

068577

P

 

, though the error from

P

H

2

.

456603 .

F

2

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