ACTIVITY 6.7
ACTIVITY 6.7
Selecting and
Rearranging
Things
OBJECTIVES
1. Determine the number of
permutations.
2. Determine the number of
combinations.
3. Recognize patterns
modeled by counting
techniques.
SELECTING AND REARRANGING THINGS
757
In working through the previous activities, you may have noticed there are many
different varieties of counting problems. Sometimes a tree or list may be feasible
to display all the possibilities. But usually the overwhelming number of possibilities makes constructing a list or tree impractical. In this activity you will explore two
types of counting problems, permutations and combinations, both of which use
the multiplication principle of counting.
Permutations are arrangements of objects, from first to last, where the order in
which the objects are selected is most important.
Example 1
You havefivetextbooks and wish to arrange them on your bookshelf in
the usual way. In how many ways can this be done?
SOLUTION
Think of the five positions on the shelf, left to right. Consider how many ways the
first space can be filled (5), then how many ways to fill the second space (4, since
there are four books remaining), and so on. You could illustrate with a tree diagram
if you wish, but a simple application of the multiplication principle will give you
5 • 4 • 3 • 2 • 1 = 120. So there are 120 different arrangements or orderings of
your five books. Equivalently, you say there are 120 permutations of any five objects.
1. Suppose you have seven different sweatshirts, and wish to wear a different one
each day of the week. In how many ways can this be done?
2. There are 26 letters in the standard English alphabet. If you consider using all
the letters to create a 26-letter word, how many different words are possible?
(You will notice this is a very large number, so you will need scientific notation to give an approximate answer.)
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CHAPTER 6
PROBABILITY MODELS
Factorial Notation
The product of any positive integer with all smaller positive integers down to
one has a special name and notation. For example, Five factorial is written as
51 = 5 . 4 . 3 . 2 - 1 = 120.
3. Use the factorial notation on your calculator (see Appendix A) to determine 12!
Appendix
Note: 0! is defined to be equal to one. (Check this on your calculator.) You can think
of it as the number of ways to arrange zero things.
Sometimes you are interested in arrangements of only some of the objects from a
larger collection. Words that are 26 letters long are a bit unrealistic, but using only
five different letters might make more sense.
Example 2
Suppose exactlyfivedifferent letters are used to make a word (in this case
a word does not have to be one found in a dictionary, but simply be a
sequence of five letters). How many different words can be made?
SOLUTION
As with the bookshelf problem, imagine five spaces, to hold each of the five letters.
But now, there are 26 choices for the first space, 25 for the second, and so on.
26 • 25 • 24 • 23 • 22 = 7,893,600 different words are possible. Notice that since
there are only five letters, this is not a full factorial. These are still called permutations. There are 7,893,600 permutations of 26 objects, taken five at a time.
Permutations of n objects taken r at a time = the first r factors of n !
Notation: nPr = n(n - 1)(« - 2)
[A]*A*,,,**,
(n - r + 1) =
(n-r)!
4. Redo Example 2 using the permutations formula above. Then, verify your
answer by using the permutation feature on your calculator (see Appendix A).
5. The governor is visiting your school and you need to assign seating for 14 students.
Unfortunately there are only six seats in the front row. In how many different ways
can you select and arrange the six lucky students?
Combinations
•
Sometimes the order in which selections are made is not important. For example,
in Problem 5, if the arrangement of the six chosen students is not important, you
would simply want to count how many ways a group of six students could be chosen
for the first row. For this revised situation you are counting the number of combinations of 14 objects taken six at a time.
ACTIVITY 6.7
SELECTING AND REARRANGING THINGS
759
Combinations are collections of objects selected from a larger collection, where
the order of selection is not important. The number of ways to select r objects
from a collection of n objects is called the number of combinations of n objects
taken r at a time.
6. In Problem 5, for each possible group of six students, how many seating
arrangements are possible?
7. If you could list all 2,162,160 permutations in Problem 5, you could also group
them according to the actual six students selected. Each group would have the
number you determined in Problem 6. Divide accordingly to determine the
number of ways to select the six students.
P
Combinations of n objects taken r at a time: nCr = !LrP
-¿r
«'
(n-r)lr\
8. Apply the combinations formula above to verify your answer to Problem 7.
9. From a student body of 230 students, how many different committees of four
students are possible? (Hint: The order in which the four students are selected
is not important.)
An interesting symmetrical property of combinations appears when you consider not
only the objects selected from a collection, but also those left behind. Does it make
sense that the number of ways to select 3 objects from a collection of 8 objects is
the same as counting the number of ways to leave 5 behind?
10. Verify the above statement by computing 8 C 3 and 8 C 5 .
Many times the most important part of the problem is determining whether or not the
order of the selections is significant. In Exercise 5 of Activity 6.2, involving ordering
a triple scoop ice cream cone, some people might consider the order of the scoops
important. Others might not care how the three flavors are placed on the cone.
11. In each situation, determine whether you are asked to determine the number of
permutations or combinations. Then do the calculation.
a. How many ways are there to pick a starting five from a basketball team of
twelve members?
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CHAPTER 6
PROBABILITY MODELS
b. How many ways are there to distribute nine different books among 15
children if no child gets more than one book?
c. How many ways are there to pick a subset of four different letters from the
26-letter alphabet?
d. How many ways can the three offices of chairman, vice chairman, and
secretary be filled from a club with 25 members?
Applications
Many variations on these basic counting techniques arise in the study of probability
and computer science. Here are a couple of examples, with problems for you
to try.
«
Example 3
How many different ten-digit binary sequences (only 0s and Is) are there
with exactly seven zeros and three ones?
SOLUTION
While the order of the binary digits is important, this counting problem is most easily solved by determining the number of combinations of ten digits taken seven at
a time. Think of the ten positions for the digits, and ask: How many ways can you
select seven positions for the zeros?
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
low ~ (7 • 6 • 5 • 4 • 3 • 2 • 1) • (3 • 2 • 1)
.10-9
3 •2
= 120.
Or equivalently, how many ways can you select three positions for the ones?
C,
IOW
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
= (3 • 2 • 1) • (7 • 6 • 5 • 4 • 3 • 2 • 1)
10-9
3 •2
= 120.
12. If a coin is flipped twenty times, how many different ways are there to get
exactly five heads?
13. What is the probability that a coin flipped twenty times will come up heads
exactly five times?
ACTIVITY 6.7
Example 4
SELECTING AND REARRANGING THINGS
761
How many ways are there to arrange the letters of the word SYSTEMS?
T S S Y MS
E
SOLUTION
The repeated S makes this more difficult than a simple permutation problem. If you
imagined the letters on tiles (like the game Scrabble), with the S tiles labeled with
numbers: S¡, S2* S3, so you can tell them apart, then there would be simply 7! ways
to arrange the seven tiles. But without the numbers, or tiles, the S's all look the
same, so you would be way over-counting the different words. Each unique word
will appear 3! times, the number of ways the three 5's can be ordered. So if you
divide out this multiplicity, you will have an accurate count:
7'
— = 7 • 6 • 5 • 4 = 840 truly unique words.
14. How many different words can be formed by rearranging all the letters of the
word MIRROR"!
SUMMARY
ACTIVITY 6.7
1. Permutations are arrangements of objects, from first to last, where the order in
which the objects are selected is most important.
2. Factorial Notation: n\ ~ n(n - l ) ( n - 2) ••• 3 • 2 • 1
3. Permutations of n objects taken r at a time = the first r factors of n !
Notation: nPr = n(n - l ) ( n - 2)
(n - r + 1) =
n\
(n - r ) !
4. Combinations are collections of objects selected from a larger collection, where
the order of selection is not important.
r
5. Combinations of n objects taken r at a time = „Cr = "-£ = 7
n'
'-r-,
r\