Q.

1: 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

Q:4.41 [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

Q: 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

Q: 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?
Solution: Prn  n! , so P312  12!  12 11 10  1320
n  r ! 9!
: The number of ways four aces can be located in a deck of 52 cards.
Solution: Apparently, order counts in this case.
52! 52!
P452    52  51  50  49  6497400 .
52  4 ! 48  !
Q:4The 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 rn  n! , so C 28  8!  8!  8  7  28
n  r ! r! 8  2!2! 6! 2! 2 1

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
 43 
 1 1  1  3 1  1 1 1
C 24 C 03 C 02 C 04 C 23 C 02 C 04 C 03 C 22  2 1 
PR  Bl  Br     
C 29 C 29 C 29 9 8
2 1
 10 5
   .27778
36 18
: 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
n!
from the 5 pennies). C rn  . So
n  r ! r!
4! 5! 43
C 24 C 05 2! 2! 5! 0! 2 1 4  3 12 1
P2 dimes         .1667
C 29 9! 9  8 9  8 72 6
2! 7! 2 1

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
20
C20  1 way to do this. We divide this by the number of ways to take 20 balls
50! 50!
from a total of 50. 50
C 20  
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
20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2 1

answer is 150  1
 2.1 10 14 . There is another way to argue this. On the
C 20 1.7619 10 13
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 20th try, when there is only one blue
ball left out of 31 remaining. If you multiply these 20 fractions together you
20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2 1 1
get   50 .
50  49  48  47  46  45  44  43  42  41  40  39  38  37  36  35  34  33  32  31 C 20

Explain intuitively C0n .

Solution: There is only one way to take nothing from n items (if order
doesn't count). C rn  n! , so
n  r ! r!
n! n!
C 0n   1.
n  0! 0! n!1

: Explain intuitively C1n .

Solution: There is n ways to take 1 item from n items, since there are n

choices. C1n  n!  nn  1!  n .

n  1!1! n  1!1!

: Explain intuitively Cnn .

Solution: There is only one way to take all from n items.
n! n! n!
C rn    1.
n  n! n! 0 ! n! 1 n!
: 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 and there are n ways to pick that one item.
n! n! n  n  1!
C nn1    n.
n  n  1! n  1! 1! n  1! 1 n  1!

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
n!
choose j items to leave out. Mathematically, C nj  and
n  j ! j!
n! n!
C nn j   . These are the same except for the order of
n  n  j ! n  j !  j ! n  j !
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, C1052 , while the numerator must be the number of ways to take 2
jacks from the four in the deck, C24  6 , multiplied by the number of ways to
take the remaining 8 cards from the 48 non-jacks in the deck, C848 . Recall
4! 48! 4  3 48  47  46  45  44  43  42  41
C 24 C848
 2 1 8  7  6  5  4  3  2 1
n!
that C rn  , so P2 jacks  
2! 2! 40!8!
. If
n  r ! r! 52
C10 52! 52  51  50  49  48  47  46  45  44  43
42!10! 10  9  8  7  6  5  4  3  2 1
we cancel the numbers 8 down through 1 and the numbers 48 down through
43 within both the numerator and the denominator, we get
6  42  41 10  9
P2 jacks 
6642
 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 .
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, C1052 , while the numerator must be the number of ways to take 2
jacks from the four in the deck, C24  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! , so
n  r ! r!
4! 4! 44! 43 44  43  42  41  40
4
C 24 C 34 C 544 2 1 5  4  3  2 1
P2 jacks, 3 tens 
2! 2! 3!1! 39!5!
  . If we cancel
52
C10 52! 52  51  50  49  48  47  46  45  44  43
42!10! 10  9  8  7  6  5  4  3  2 1
the 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
2 1 6  4  42  41  40 10  9  7  6
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
52  51  5  49  2  47  46 2809475760
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  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
Then PH 2  C2 C523  78 9139   712842  .274280 and PF 2  C2 C523
13 39 12 40

C5 2598960  2598960 C5

66 9880   652080  .250900 . The hard part is finding the intersection of these
2598960  2598960
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.

(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 cards. The probability is thus
PF 2  H 2 
178230
 .068577 . Incidentally, we could not get this by multiplying
2598960
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   .250027  PF 2 , though the error from assuming independence
178230
712842
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 .