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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 3rd ed) 19.
3. Combinations.
4.48 [4.45*], D&C pg. 97 20-21, 29-34. I1-I4.
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Exercise 4.38 [4.36 in 9th](Not in 8th 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: 310  59049
Exercise 4.41 [4.39 in 9th](Not in 8th 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 9th] (Not in 8th 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 9th] (Not in 8th edition): The Big Triple consists of picking the correct order of
finish of the first 3 horses in the 9th race. If there are 12 horses entered in the race, how many outcomes are
there?
n!
12!
 12 11 10  1320
Solution: Prn 
, so P312 
9!
n  r !
Downing and Clark, pg. 85, Application 19: The number of ways four aces can be located in a deck of 52
cards..
52!
52!
Solution: Apparently, order counts in this case. P452 

 52  51  50  49  6497400 .
52  4 ! 48  !
Exercise 4.48 [4.45 in 9th] (Not in 8th 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?
n!
8!
8!
87


 28
Solution: C rn 
, so C 28 
n  r ! r!
8  2!2! 6! 2! 2 1
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  

10
5

 .27778
36 18
C 24 C 03 C 02
C 29

C 04 C 23 C 02
C 29

C 04 C 03 C 22
C 29
 43

11  1 3 1  111

2

1



9 8
2 1
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 29 ways to pick 2 coins from 9, but only
C 24 ways to pick 2 dimes from 4 (and nothing from the 5 pennies). C rn 
P2 dimes  
C 24 C 05
C 29
n!
n  r ! r!
. So
4! 5!
43
2! 2! 5! 0! 2 1 4  3 12 1




  .1667
9!
9  8 9  8 72 6
2 1
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?
20
Solution: There are 20 blue balls and we need all of them. There is only C20
 1 way to do this. We divide
50
this by the number of ways to take 20 balls from a total of 50. C 20

50!
50!

50  20 ! 20! 30 ! 20!
50  49  48  47  46  45  44  43  42  41  40  39  38  37  36  35  34  33  32  31
 1.7619 10 13 . So the answer is
20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2 1
1
1

 2.1 10 14 . There is another way to argue this. On the first try the chance of getting a
50
13
C 20 1.7619 10

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
because there are 19 blue balls left among the 49 remaining balls. This
49
goes on until the 20 try, when there is only one blue ball left out of 31 remaining. If you multiply these 20
fractions together you get
20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2 1
1

 50 .
50  49  48  47  46  45  44  43  42  41  40  39  38  37  36  35  34  33  32  31 C 20
th
Downing and Clark, pg. 85, Application 30: Explain intuitively C0n .
Solution: There is only one way to take nothing from n items (if order doesn't count). C rn 
C 0n 
n!
n!

 1.
n  0! 0! n!1
Downing and Clark, pg. 85, Application 31: Explain intuitively C1n .
Solution: There is n ways to take 1 item from n items, since there are n choices.
n!
nn  1!
C1n 

 n.
n  1!1! n  1!1!
Downing and Clark, pg. 85, Application 32: Explain intuitively C nn .
Solution: There is only one way to take all from n items. C rn 
n!
n  n! n!

n!
n!

1.
0 ! n! 1 n!
n!
n  r ! r!
, so
251solnI 10/09/03
Downing and Clark, pg. 85, Application 33: Explain intuitively C nn1 .
Solution: There are n ways to take n  1 items from n items. This is because we must leave out one item
n!
n!
n  n  1!
and there are n ways to pick that one item. C nn1 


n.
n  n  1! n  1! 1! n  1! 1 n  1!
Downing and Clark, pg. 85, Application 34: Compare C nj and C nn j . Explain intuitively.
Solution: The number of ways to take j items from n items must be the same as the number of ways to
take n  j items from n 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 nj 
C nn j 
n!
and
n  j ! j!
n!
n!
. These are the same except for the order of computation.

n  n  j ! n  j !  j ! n  j !
Problem I1: What is the probability of two jacks in a hand of ten cards?
52
Solution: The denominator must be the number of ways to take 10 cards from 52, C10
, while the
numerator must be the number of ways to take 2 jacks from the four in the deck, C 24  6 , multiplied by the
number of ways to take the remaining 8 cards from the 48 non-jacks in the deck, C848 . Recall that
4! 48!
4  3 48  47  46  45  44  43  42  41
4 48
C
C
n
!
2
!
2
!
40
!
8
!
8  7  6  5  4  3  2 1
, so P2 jacks  2 528 
. If we
C rn 
 2 1
52!
52  51  50  49  48  47  46  45  44  43
n  r ! r!
C10
10  9  8  7  6  5  4  3  2 1
42!10!
cancel the numbers 8 down through 1 and the numbers 48 down through 43 within both the numerator and
6  42  41
10  9
6642
the denominator, we get P2 jacks 
 6  42  41

 .1431 .
52  51  50  49
52  51  50  49 46410
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?
52
Solution: The denominator must be the number of ways to take 10 cards from 52, C10
, while the
numerator must be the number of ways to take 2 jacks from the four in the deck, C 24  6 , multiplied by the
number of ways to take 3 tens from the four in the deck, C34  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, C544 . Recall that
C rn 
n!
n  r ! r!
, so
4! 4! 44!
43
44  43  42  41  40
4
2! 2! 3!1! 39!5!
2

1
5  4  3  2 1
. If we cancel the
P2 jacks, 3 tens 


52!
52  51  50  49  48  47  46  45  44  43
10  9  8  7  6  5  4  3  2 1
42!10!
numbers 5 down through 1 and the numbers 44 down through 43 within both the numerator and the
denominator, we get
43
 4  42  41  40
6  4  42  41  40 10  9  7  6
2 1
P2 jacks, 3 tens 

52  51  50  49  48  47  46  45 52  51  50  49  48  47  46  45
10  9  8  7  6
42  41  8  7  6
564816


 .00021 The last line involves a great deal of canceling, so I urge
52  51  5  49  2  47  46 2809475760
you to check my work and see what you get.
C 24 C 34 C 544
52
C10
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  PF 2  P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.
30 cards that are neither face cards or hearts
C 13C 39 78 9139  712842
C 12C 40
Then PH 2  2 523 

 .274280 and PF 2  2 523
2598960  2598960
C5
C5

66 9880   652080
2598960  2598960
 .250900 . 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
(a) Two nonface hearts, two nonheart face cards, no heart face cards. This can happen in
C210C29C03C130  4536130  48600 ways.
.
(b) One nonface hearts, one nonheart face cards, one heart face cards. This can happen in
C110C19C13C230  1093435  117450 ways.
(c) No nonface hearts, no nonheart face cards, two heart face cards. This can happen in
C010C09C23C330  1134060  12180 ways.
We can conclude that there are 48600+117450+12180=178230 ways to get both 2 hearts and two face
178230
cards. The probability is thus PF 2  H 2 
 .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
178230
independent as is demonstrated by PF 2 H 2 
 .250027  PF 2 , though the error from
712842
assuming independence would be small.
We thus conclude that PH 2  F 2
.  PH 2  PF 2  PH 2  F 2  .274280  .250900  .068577  .456603 .