Counting Techniques
In probability and statistics, we sometimes need to count the number of ways that
a phenomenon can occur. Possible examples are the number of license plates of
a certain form, the number of different ways ten people can line up in a row, the
number of different ways ten people can order themselves in a circle, or the
number of different ways to pick three people out of ten to form a team or
committee. To solve problems like these we often use general counting
techniques.
Fundamental Principle of Counting (FPC)
Example 1 Let us conduct a random experiment during which we first flip a coin
and then roll a die. For example, one of the outcomes of the random experiment is
getting a head (H) from flipping the coin and then getting a “1” from rolling the die.
We want to count all of the different ways in which the random experiment can
possibly occur. Since this random experiment consists of a sequence of actions, it is
useful to draw a tree diagram, which is a systematic way of making sure we
account for all the outcomes in this experiment. At the beginning of the random
experiment, we imagine ourselves at the root of the tree. We then move along the
branches of the tree as the random experiment occurs. It is convenient to have the
tree lay horizontally. From the root of the tree, we will first branch to all the different
outcomes of the first phenomenon, the coin toss (outcome head or tail), and then
branch to all the possible outcomes of the second phenomenon, the die roll
(outcome 1, 2, 3, 4, 5, or 6). We get the following tree diagram:
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Each complete path through the tree is a particular outcome of the random
experiment. The second path from the top indicates that the coin is a head and
the die is a two. If one counts the leaves at the very end of the tree, then we find
that this random experiment could occurs in twelve different ways. All the possible
outcomes in this experiment of tossing a coin followed by rolling a die are listed
using set notation below:
S = {(H,1), (H.2), (H.3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
S represents the sample space of this experiment, which is the set of all possible
outcomes of the random experiment. We sometimes use ordered pair notation to
list elements of a sample indicating the elements represent ordered arrangements.
Alternatively, we could also count the paths using the Fundamental Principle of
Counting. Since each of the two outcomes of the coin flip is followed by the six
outcomes of the die roll, there are a total of 2  6  12 ways the random experiment
could occur.
It is important to note that in Example 1, the experiment consists of sequence of
random experiments that are ordered. First, we flipped a coin and then we rolled a
die. This produced outcomes that are ordered arrangements, and the
Fundamental Principle counts the number of ordered arrangements. The
Fundamental Principle of Counting is generalized as below.
The Fundamental Principle of Counting (FPC) says that if one phenomenon occurs
in r ways and each of the r ways in which this phenomenon occurs is followed by a
second phenomenon occurring in s ways, then the number of ways both
phenomena together occur is r  s ways ( r  s ordered arrangements). Furthermore,
if each of the r  s ways in which the first two phenomena occur is followed by a
third phenomenon occurring in t ways, then the number of ways all three
phenomena occur together is r  s  t ways ( r  s  t ordered arrangements). Similarly
we could add a forth, fifth, or any other number of phenomena and obtain similar
results.
Example 2 Let us consider another problem where one may use a tree diagram
and the FPC to solve the problem. How many ways can we arrange in order any
three of the four letters a, b, c, and d if we cannot use a letter more than once?
For example, here are three possible outcomes: abc, abd, cba but, we cannot
have aba, aab or bba since we cannot use a letter more than once. Let us draw a
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tree.
Let us write down all the outcomes of this experiment in the set notation below.
S = { abc, abd, acb, acd, adb, adc, bac, bad, bca, bcd, bda, bdc, cab, cad,
cba, cbd, cda, cdb, dab, dab, dba, dbc, dca, dcb}
Now, let us discuss the set-up of this tree-diagram. We first branch to each of the
four letters that could be the first letter in the arrangement. Next, for each of these
four letters, we branch to any of the letters that could be the second letter. Finally,
from each of the second possible letters, we branch to any of the letters that could
be third in our ordered arrangement.
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From each of the four first tiers of branches follows only three branches because
the particular letter selected first cannot be the second letter. There are only two
branches from each of the second tier of branches because the first and second
letters selected cannot be the third letter. The first path through the tree represents
the arrangement abc and the third path acb, which are counted as different. In
other words, trees count ordered arrangements. If we reorder the same three
letters obtaining a different order, then the tree counts these two arrangements as
different.
Another way to count the total number of outcomes produced in this experiment
without writing out all the outcomes or using the tree-diagram is to use the FPC. By
the FPC, there are a total of 4  3  2  24 ordered arrangements of three of the four
letters. Here is the reason. There are four possible ways to pick a letter for the first
tier of the tree-diagram. Once a letter is selected for the first tier, we cannot use
that letter again, which leaves us with only three possible ways to pick a letter for
the second tier. Once the letters are used for the first and second tiers, we cannot
use these two particular letters again; therefore, we only have two possible ways to
pick a letter for the third tier.
Example 3 At a restaurant, for a fixed price a patron may choose from one of four
salads, one of five entrées, and one desert. How many different meals are possible,
if the patron must select one salad, one entrée, and one dessert?
Let us draw a tree.
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One possible meal a patron may have is the first path through the tree, “S1E1D”. By
the FPC, the total number of possible meals is 4  5 1  20.
If one multiplies various natural numbers together to get the answer to a counting
problem, then the problem is using the Fundamental Principle of Counting.
Underlying every Fundamental Principle of Counting problem is a tree, which one
could draw or at least imagine. However, some problems have so many
outcomes, it is not feasible to draw a tree diagram but at least one should try to
imagine a tree-diagram or perhaps sketch enough of the tree so that one is sure
that the multiplications one is performing are correct.
Example 4 There are 5 children in a preschool. How many different ways can the
teacher sit these 5 children in a row of 5 chairs?
For chair #1, the teacher can seat any one of the 5 children.
For chair #2, the teacher can seat any one of the 4 remaining children.
For chair #3, the teacher can seat any one of the 3 remaining children.
For chair #4, the teacher can seat either one of the 2 remaining children.
For chair #5, the last chair, the teacher has only 1 child to seat.
Using the FPC, the total number of different ways the teacher can seat 5 children in
a row of 5 chairs is 5  4  3  2 1  120 .
This example brings us to a very useful mathematical notation, called factorial.
Factorial Notation
Sometimes the following notation is very useful. If n is a whole number, which means
n is an element of the set {0, 1, 2, 3, 4, 5, 6 ...}, then n factorial, written n!, is defined
0!  1;1!  1; 2!  2 1;3!  3  2 1; 4!  4  3  2 1;... , etc. For example,
10!  10  9  8  7  6  5  4  3  2 1  3, 628,800.
In general, the pattern for n! is
n !  n(n  1)(n  2)(n  3)  4  3  2 1.
The reason 0! =1 is because 0! is the number of ways nothing can happen, which is
1. 1! = 1 also because 1! is the number of ways one thing can happen. We shall
see that factorial notation is sometimes very useful in counting.
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Permutations
A permutation is an ordered arrangement of r distinct objects, from a set of n
different objects. The size of a permutation is the number of distinct objects in the
arrangement.
In example 2 above, each of the arrangements is an ordered arrangement of
three of the four distinct letters a, b, c, and d. To have a permutation the following
must be true.
1. The objects making up the permutation must all be different. We did not
count aba or any arrangement in which one or more letters is the same. All
of the arrangements consisted of three distinct letters.
2. The arrangements must be ordered. For example, we counted each of the
arrangements of the letters a, b, & c as different. The permutation abc is not
the same as cba because the order of the three letters is different. All
possible permutations of the letters a, b, & c are the six different ordered
arrangements of these three letters: abc, acb, bac, bca, cab, & cba.
By definition, we must construct a permutation by choosing the r objects from the n
distinct objects without replacement. Furthermore, in a permutation, the objects
are ordered.
P to represent the total number of different permutations of
n r
size r that can be selected from n distinct objects. In example 2 above, we
computed,
4 P3 = 4  3  2  24 ,
We use the notation
the total number of permutations of size three which can be selected from the four
distinct letters, a, b, c, & d.
n
Pr 
n!
, 0  r  n , r and n whole numbers
(n  r )!
is the general formula for computing the total number of permutations of size r
selected from n distinct objects.
Example 5 Let us look at an example that will indicate why the permutation
formula is true. Similar to example 2 above, let us compute how many ways we
can arrange in order any three of the first 8 letters of the alphabet a, b, c, d, e, f, g,
h. Here we are finding 8 P3 , the number of permutations of size 3 selected from the
first 8 letters of the alphabet. Similar to example 2, 8 P3  8  7  6  336 since the first
letter can be any of the 8 letters, the second letter can be any of the seven letters
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left, and the third letter can be any of the six letters left. The permutation formula
gives the same result:
P 
8 3
8!
8! 8  7  6  5  4  3  2 1
 
 8  7  6  336 .
(8  3)! 5!
5  4  3  2 1
The 5! in the denominator above cancels with all but the first three factors of 8!
leaving 8  7  6.
Example 6 Consider a bag of billiard balls containing two identical red balls, a
blue ball, and a yellow ball. Select each of the balls one at a time and arrange
them in the order selected. Using this method, how many different ways can one
select and arrange all four balls?
This is not a permutation problem because in a permutation problem, all of the
objects must be distinguishable and the two red balls are identical. Nonetheless,
we can solve this problem using permutations by pretending that the two red balls
are distinguishable. If the two red balls are distinguishable, then the number of
4!
permutations of size four that can be selected from the four balls is 4 P4 
 24 ,
(4  4)!
which measures the number of the ordered arrangements of the four balls. Twentyfour, however, is not quite correct because two of the balls are not distinct.
Suppose a particular ordered selection is a yellow, a red, a blue, and a red ball.
Since actually the two red balls are indistinguishable, if the two red balls were
permuted (switched), we do not get a different outcome. The 24 permutations
counted above, however, does count the permuting of the two red balls as
different. The permuting of the two red balls occurs in 2! ways. Thus, 4 P4  24 is 2!
times too large. With the two red balls identical the number of different ways one
4!
P (4  4)!
can select the balls is 4 4 
 12 .
2!
2!
Combinations
A combination is an unordered group of r distinct objects, from a set of n different
objects. The size of a combination is the number of distinct objects in the
unordered group. We use the notation n Cr to represent the total number of
different combinations of size r that can be selected from n distinct objects.
n
Cr 
n!
, 0  r  n , r and n whole numbers
(n  r )!r !
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is the general formula for computing the number of combinations of size r selected
from n distinct objects.
Similar to permutations, combinations require that we choose the r objects from the
n distinct objects without replacement.
The general formula for the number of combinations may be derived using the
permutation formula. The next example outlines the argument.
Example 7 How many different unordered groups of any three of the four letters a,
b, c, and d exist?
The symbol for the answer is n Cr  4 C3 , which is the number of combinations of size
r = 3 that can be selected from the n = 4 letters a, b, c, & d.
Let us list all such unordered groups and for each unordered group list all the
permutations that can be made from the letters in that group.
1. Unordered group a, b, c, from which 3! = 6 permutations can be constructed,
namely abc, acb, bac, bca, cab, & cba.
2. Unordered group a, b, d, from which 3! = 6 permutations can be
constructed, namely abd, adb, bad, bda, dab, & dba.
3. Unordered group a, c, d, from which 3! = 6 permutations can be constructed,
namely acd, adc, cad, cda, dac, & dca.
4. Unordered group b, c, d, from which 3! = 6 permutations can be constructed,
namely bcd, bdc, cbd, cdb, dbc, & dcb.
Therefore, 4 C3  4 . Notice that all 24 permutations of any three of the four letters a,
b, c, and d are listed once and only once among the four groups above. Thus, we
have 4  6  24 , or 4 C3  3!  4 P3 . With n = 4 and r = 3 dividing by r! = 3!, we have
4!
4!
n!
(4  1)!
4 P3



.
n Cr  4 C3 
3!
3!
(4  1)! 3! ( n  r )! r !
The general relationship for the number of permutations and combinations of size r
that can be selected from n distinct objects is
n
Cr  r !  n Pr
because if one finds all the unordered groups of size r which can be selected from
n distinct objects, n Cr , and then finds all the r! permutations of the r objects in each
unordered group, one gets all the permutations of size r of the n objects.
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By dividing both sides of the above equation by r! and then using the general
permutation formula, we get
n!
n!
(n  r )!
n Pr


.
n Cr 
r!
r!
(n  r )! r !
Example 8 From a class of 16 students, 4 will be chosen to do a job. How many
different groups of students are possible?
This is a combination problem because the 16 students are distinct and there is no
order among the students associated with doing the job. The number of different
groups of students possibly doing the job is
16
C4 
16!
16!
16 15 14 13 12! 16 15 14 13



 1820.
(16  4)! 4! 12! 4!
12! 4!
4  3  2 1
Notice how the 12! cancels into the 16! to help simplify the calculation.
Example 9 From a class of 16 students, how many ways could the jobs of
President, Vice President, Secretary, and Treasurer of the class be filled by the
students?
This is a permutation problem because the 16 students are distinct and there is an
order associated with filling the four class officers. If four particular students hold
the offices and we permute (switch) these students among the offices, we get a
different arrangement. The number of different arrangements of students holding
the four offices is
16
P4 
16!
 43680.
(16  4)!
Example 10 Using an ordinary deck of cards with 52 cards, how many different 5card hands have four of a kind and one other card?
An example of such a hand is the queen of hearts, queen of diamonds, queen of
clubs, queen of spades, and the eight of diamonds. Card hands like this are
combination problems because reordering the cards in the hand does not yield a
different hand. First, let us select the value of the card of which we will have four of
13!
13!
a kind. The number of ways this can happen is 13 C1 

 13 ways
(13  1)!1! 12!
because the value the card of is any one of the thirteen different values of cards in
the deck: ace, king, queen, jack, ten, nine, eight, seven, six, five, four, three, or
Page 9
two. Next, let us compute the number of combinations of the particular four cards
4!
with the value that we just selected, which occurs in 4 C4 
 1 way, as we
(4  4)! 4!
need all four of these cards in the hand. Finally, Let us choose the fifth card, which
can be any one of the forty-eight cards remaining, which occurs in
48!
 48 ways.
48 C1 
(48  1)!1!
The number of different 5-card hands that have four of a kind and one other card is
13 C1  4 C4  48 C1  13 1 48  624 . This is because for each of the 13 ways we can select
the value of the card of which we can have four of a kind is followed by the one
way in which we can combine these fours cards in an unordered hand and in turn
follows the 48 ways we may select the remaining card. In other words the answer
of 13 1 48 is because of the Fundamental Principle of Counting . The following tree
illustrates the Fundamental Principle of Counting used to solve this problem.
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The paths ending in a dot represent missing or incomplete paths. From each of the
thirteen paths in the second tier branch 48 paths, most of which are missing in the
tree diagram. If you look at the end leaf, the three of spades, the hand
represented by this particular path consists of four twos and the three of spades. In
the full tree each five card hand with four of a kind is listed once and only once in
the tree. For many Fundamental Principle of Counting problems, the tree behind
the multiplication has far too many paths to draw. For such problems, it is
sometimes helpful to draw the main branching tiers and some of the paths of the
tree or to imagine the tree in your mind.
Probabilities and Counting Techniques
The counting rules can be combined with the probability rules to solve many types
of probability problems. We will use the counting techniques to count all the
possible outcomes for experiments and for events. Recall that if the sample space
is equally likely, the probability of an event E is
P( E ) 
E
S
where |E| is the number of elements in E and |S| is the number of elements in the
sample space. Note that when E is a set and|E| is not the absolute value of E, but
rather is the number of elements in E.
Example 11 Let us roll a fair die. The sample space is
S = {1, 2, 3, 4, 5, 6},
the set of all the outcome of the die. |S| = 6 because there are six elements in S.
Let E be event of the die is a one or a two. Therefore, E = {1, 2} and |E| = 2.
Because the die is fair, the sample is equally likely and, therefore
P( E ) 
E
S

2 1
 .
6 3
Example 12 What is the probability of getting 4 aces when 5 cards are drawn
randomly from an ordinary deck of 52 cards?
As in example 10, we will consider unordered hands of five cards. In this case, the
order of the 5 cards drawn does not matter. The sample space, S, is the set of all
unordered five card hands. Thus,
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S  52 C5  2,598,960.
Let A be the event of getting 4 aces when 5 cards are drawn.
There is only 4 C4  1 way to get 4 aces and then there are
48
C1  48 ways to get the
other card. Therefore, using the FPC, the total number of outcomes in event E is
4 C4  48 C1  1 48. Notice because the sample space is equally likely, the number of
elements in an event and the number of ways the event can occur is the same.
P(4 aces & one other card) = P(A) =
A

S
4
C4  48 C1
48

 0.0000185
2,598,960
52 C5
Example 13 In a parent-teacher association, there are 7 parents and 5 teachers. A
committee of 5 members is to be chosen.
a) What is the probability that the committee has 3 parents and 2 teachers?
b) What is the probability that the committee has more than 3 parents?
First, there are 12 different members and we would like to select 5 to form a
committee. For a committee, the order of the outcomes does not matter.
Therefore, the total number of ways to select 5 members out of 12 is S  12 C5  792 .
This is the total number of outcomes in the sample space.
a) Let A be the event of getting 3 parents and 2 teachers in a committee of 5
members. The number of outcomes in event A can be counted by
A  7 C3  5 C2  35 10  350 . There are 7 C3  35 ways of selecting 3 parents out of 7,
and then there are 5 C2  10 ways of selecting 2 teachers out of 5. By the FPC, the
total number ways of selecting 3 parents and 2 teachers is 35 10  350 .
P(3 parents and 2 teachers) = P(A) =
A
S

7
C3  5 C2 350

 0.442  44.2%
792
12 C5
b) Let B be the event of getting more than 3 parents in a committee of 5 members.
To count the total number of outcomes in event B, we will need to count all the
outcomes with 4 parents and all the outcomes with 5 parents.
|B| = number of outcomes with 4 parents + number of outcomes with 5 parents
= 7 C4  5 C1 + 7 C5  5 C0 = 175 + 21 = 196.
P(more than 3 parents) = P ( B ) 
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B
S

196
 0.247  24.7%.
792
Example 14 Suppose a photographer randomly arranges 6 students, including
Maya and Elena, in a row of 6 chairs for a class picture. What is the probability that
Maya and Elena will be sitting next to each other in the class picture?
The total number of ways the photographer can randomly arrange 6 students in a
row of 6 chairs is S  6!  720 . Note that since the students will be ordered, we have
ordered arrangements. So, we cannot use combinations for this problem.
Let A be the event of Maya and Elena sitting next to other. We would like to count
all the different ways of arranging 6 students, but keeping Maya and Elena right
next to each other.
If Maya sat on chair #1 and Elena sat on chair #2, then there are 4 students to
choose from for chair #3, etc…
Maya
chair #1
Elena
chair #2
4 choices
3 choices
2 choices
1 choice
chair #3
chair #4
chair #5
chair #6
The total number of ways to have the above arrangement is 11 4  3  2 1  4!. This is if
Maya sat on chair #1 and Elena sat on chair #2. But Maya can sit on chair #2 and
Elena can sit on chair #1, and this will give us another 4! = 24 arrangements; or
Maya can sit on chair #2 and Elena sit on chair #3, which will result in another 4! =
24 arrangements. We can systematically rotate Maya and Elena down and count
all the possible line-ups. Continuing in this way, we can see there are 10 ways to
place Maya and Elena next to each other. For each of these, there are 4! ways to
place the remaining 4 students. So, there is a total of 10  4!  240 different ways of
arranging 6 students in a row of 5 chairs with Maya and Elena sitting next to each
other (this is the number of outcomes in event A).
Another approach to count the total number of outcomes in event A is to pretend
Maya and Elena is one person for now. And then consider how many ways can we
arrange 5 students in 5 chairs. The answer is 5!  120 . Since there are two ( 2!  2 )
different ways to rearrange Maya and Elena, we have 2!5!  240 .
P(Maya and Elena sitting together) =
A
S
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
240 1
  0.333  33.3%
720 3
Exercises
1. Find 7!
2. Find 0!
3. Find 5! 6!
4. Find
20!
.
16!
5. Find
20!
.
16! 4!
6. Find
12
C8 .
7. Find
12
P8 .
8. Find
12
C0 .
9. Find
12
P0 .
10. A person is selecting two blocks from five blocks labeled A, B, C, D, and E.
a. How many permutations are possible? Draw a complete tree for the
permutations. List all of the permutations.
b. How many combinations are possible? List all of the combinations.
11. From nine people, how many groups of six people are possible?
12. Determine if each is a permutation of a combination:
a. A committee with three people.
b. Three people serving as the chair, secretary, and treasure of the
committee.
c. Three students standing in a line.
13. Draw a tree for find all the permutations of the letters xyz. How many
different permutations of the three letters are there?
14. Find all the ordered groups of size 6 that can be selected from 9 distinct
objects.
15. Find all the unordered groups of size 0 that can be selected from 5 distinct
objects. Explain what your answer measures.
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16. Find all the ordered groups of size 0 that can be selected from 5 distinct
objects.
17. Find all the unordered groups of size 1 that can be selected from 7 distinct
objects.
18. Find all the ordered groups of size 1 that can be selected from 7 distinct
objects.
19. How many combinations of the digits 0,1, 2, 3, 4, 5, 6, 7, 8, and 9 of size 3 are
there?
20. How many three-digit natural numbers are there?
21. Let us conduct a random experiment in we first flip a coin, then roll a foursided die, with sides numbered 1, 2, 3, & 4, and then flip another coin. Draw
a tree for the sequence of outcomes of the random experiment. In how
many different ways can the random experiment possible occur.
22. Consider a bag of billiard balls containing three identical red balls, a blue
ball, and a yellow ball. Select each of the balls one at a time and arrange
them in the order selected. How many different ways can this occur?
23. How many different lower case, four-letter code words can be formed using
the twenty-six letters of the alphabet?
24. How many different four-letter code words can be formed using the twentysix letters of the alphabet if both upper and lower case letters are used?
25. At a restaurant, for a fixed price a patron must choose from one of five
appetizers, one of six salads, one of five entrées, and one of two deserts with
no substitutions allowed How many different meals are possible?
26. From the seventeen employees and the owner of a business, the owner
along with three employees will be chosen to be members of a four-person
evaluation committee. How many different evaluation committees could be
formed?
27. From a group of twelve people on a committee, how many ways could the
jobs of Chair and Recording Sectary be filled by the members of the
committee?
28. Six different books are on a shelf. In how many different ways could you
arrange them on the shelf?
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29. Consider the word earth.
a. Using all the letters in the word, how many permutations are there?
b. Using all the letters in the word, how many permutations are there with
the letters t and h next to each other?
c. Using all the letters in the word, how many permutations are there with
the letters e, a, and r together?
d. Using all the letters in the word, how many permutations are there with
first letter e?
30. How many different ordered arrangements are there of the four letters in the
word moon?
31. How many different ordered arrangements are there of the eight letters in
the word moonbeam?
32. There are five balls identical except for the color of the balls in a bag. There
is a red, a white, a blue, a green, and a yellow ball.
a. How many different ways can one select one ball?
b. How many different ways can one select two of the balls at the same
time?
c. How many ways can one select three of the balls at the same time?
d. How many ways can one select four of the balls at the same time?
e. How many ways can one select all five of the balls at the same time?
33. There are five balls in a bag. There are two red, and three white balls. The
two red balls are identical and the three white balls are identical.
a. How many ways can a person select one ball?
b. How many ways can a person select two of the balls at the same
time?
34. A worker is packing small boxes into a larger box. There are twelve smaller
boxes each of a different color. If the worker puts exactly eight of the small
boxes into the larger box and must make sure all the boxes in the larger box
are a different color, how many different ways are there to pack the larger
box?
35. A worker is packing small boxes into a larger box. There are twelve different
colors of small boxes and an unlimited number of each color. If the worker
puts exactly eight of the small boxes into the larger box and may put more
than one box of the small color into the larger box, how many different ways
are there to pack the larger box?
36. There are five freshman, six sophomores, eight juniors, and fourteen seniors in
a Campus Club. The club is to send five representatives to a Conference.
a. How many different groups of people may be sent to the conference?
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b. If one freshman, one sophomore, one junior, and two seniors must
attend, how many difference groups of people may be sent to the
conference?
37. A basketball coach must choose a team five players from his fifteen
players. How many different teams can the coach choose if each player
can play each position?
38. A basketball coach must choose a team five players from his fifteen
players. How many different teams can the coach choose if the team
consists of one center, two guards, and two forwards? Assume two of the
fifteen players play center position; seven other players play the guard
position and the remaining six players play the forward position.
39. Using an ordinary deck of cards with 52 cards, how many possible five-card
hands are there?
40. Using an ordinary deck of cards with 52 cards, how many possible five-card
hands are there which contain exactly two pairs? For example, a hand with
two kings, two threes, and a queen is one such hand.
41. Using an ordinary deck of cards with 52 cards, how many possible five-card
hands are there which contain a full house, which is a hand with three of a
kind and a pair? For example, a hand with three kings (three of a kind) and
two queens (a pair) is one such hand.
42. Using an ordinary deck of cards with 52 cards, how many possible five-card
hands are there which contain four of sevens?
43. Using an ordinary deck of cards with 52 cards, how many possible five-card
hands are there in which every card is a heart?
44. How many different ordered arrangements are there which use the all of the
eleven letters in the word Mississippi?
45. How many different ways can an eight-question true-false test be answered?
46. How many different ways can an eight-question multiple-choice test be
answered if each question has five possible answers?
47. How many different ways can an eight-question test be answered if there are
four true-false questions and four multiple choice questions each of which
has five possible answers?
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48. How many possible license plates are there of the form 5ADF6124?
Specifically, first letter is any digit except for zero, the next three characters
are each any letters of the alphabet, and the next four digits are any of the
digits.
49. How many possible license plates are there of the form 5ADF6124 if no
license plate can end in four zeros? Specifically, first letter is any digit except
for zero, the next three characters are each any letters of the alphabet, and
the next four digits are any of the digits except not all four digits are zero. For
example, the license plate 5ADF0000 is not possible, but license plate
5ADF0910 is possible.
50. How many different ways can 10 people can line up in a line?
51. How many different ways can 10 people can stand in a circle?
52. There are seven women and eight men working for a company. From this
group a committee of six is to be chosen.
a. How many possible committees are there?
b. How many possible committees are there which consist of all women?
53. There are seven women and eight men working for a company. From this
group a committee of six is to be chosen with three women and three
men. How many possible such committees are there?
54. A box contains 24 transistors, 4 of which are defective. If 4 are sold at
random, find the following probabilities:
a. Exactly 2 are defective.
b. None is defective.
c. All are defective.
d. At least one is defective.
55. Find the probability of selecting 3 English books and 4 math books from 8
English books and 9 math books if the books are selected at random.
56. The composition of the Senate is 49 Republicans, 50 Democrats, and 1
Independent. A new committee is being formed to study ways to benefit the
arts in education. If 3 Senators are selected at random to form the
committee, the find the following probabilities:
a. What is the probability that they will all be Republicans?
b. What is the probability that they will all be Democrats?
c. What is the probability that there will be 1 from each party, including the
Independent?
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57. A monkey is trained to arrange wooden blocks in a straight line. He is given 6
wooden blocks showing the letters A, E, H, L, M, and T.
a. What is the probability that the monkey will arrange the letters in a line to spell
the word HAMLET?
b. What is the probability that the arrangement will end in a vowel?
c. What is the probability that the two vowels are next to each other?
58. In California’s Super Lotto Plus lottery game, winning the jackpot requires that you
select the correct five numbers between 1 and 47 inclusive and, in a separate
drawing, you must also select the correct single number between 1 and 27 inclusive.
Find the probability of winning the jackpot.
59. The ice cream franchise Baskin-Robbins advertises thirty-one different flavors
of ice cream. Suppose you order a triple-scoop cone and tell the vendor to
pick three different flavors at random. Find the probability that you get:
a. the three flavors vanilla, chocolate, and strawberry.
b. chocolate as one of the flavors.
60. Amy, Bill, Cindy, and Dan are invited to a party. If they arrive at different
times and at random, find the probability that:
a. they will arrive in the order Amy, Bill, Cindy, and Dan.
b. Amy will arrive first.
c. Amy will arrive first and Dan last.
61. In a group, there are three teachers, two lawyers, and three doctors. If they
are seated in a row, find the probability that those of the same profession sit
together.
62. From a group of ten teachers, fifteen students, and twelve district officials, a
committee of six is selected at random. What is the probability that the
committee consists of three teachers, two students, and one district official?
63. An urn contains 8 chips, numbered 1 through 8. A sample of 3 is drawn
without replacement. What is the probability that the largest chip in the
sample is a “5”?
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