Transcript 5.1 General Counting Methods for Arrangements and Selections

5.1 General Counting Methods for
Arrangements and Selections
Tucker, Applied Combinatorics, Sec. 5.1, Patti Bodkin
and Tamsen Hunter
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Section 5.1
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Principals
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The Addition Principal
If there are r1 different objects in the first set, r2
objects in the second set, . . . , and the rm objects in
the mth set, and if the different sets are disjoint,
then then number of the ways to select an object
from one of the m sets is r1  r2  ...  rm .
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Principals

The Multiplication Principal
Suppose a procedure can be broken into m
successive (ordered) stages, with r1 outcomes
in the first stage, r2 outcomes in the second stage,
. . . , and rm outcomes in the mth stage. If the
number of outcomes at each stage is independent of
the choices in previous stages and if the composite
outcomes are all distinct, then the total procedure
has r1  r2 ...  rm different composite outcomes.
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Example 1
There are:
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five different Spanish books
six different French books
eight different Transylvanian books.
How many books are there to pick an (unordered) pair of
two books not both in the same language?
If one Spanish and one French book are chosen, using the multiplication
principal, the selection can be done; 5X6=30.
If one Transylvanian book and one Spanish book, then 5X8=40.
If one Transylvanian and one French book, then 6X8=48 ways.
These three types of selections are disjoint , so by the addition principal there
are 30+40+48=118 ways in all.
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Tip for Solving Problems
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The preceding example typifies a basic way of thinking in
combinatorial problem solving: Always try first to break a
problem into moderate number of manageable subproblems.
There may be cleverer ways, but if we can reduce the original
problem to subproblems with which we are familiar, then we are
less likely to make a mistake.
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Example 2
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How many ways are there to form a three-letter
sequence using the letters a, b, c, d, e, f:
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1: with repetition of letters allowed?
2: without repetition of any letter?
3: without repetition and containing the letter e?
4: with repetition and containing the letter e?
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Example 2.1:
How many 3-letter sequences with repetition allowed?
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With repetition, we have six choices for each letter
in the sequence. So by the multiplication principle
there are 6 * 6 * 6  216 three-letter sequences
without repetition.
a
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a
b
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Example 2.2:
How many 3-letter sequences without repetition?
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Without repetition, there are six choices for the
first letter. For the second letter, there are five
choices. Similarly for the third letter, there are
four choices. Thus there are 6*5* 4  120 threeletter sequences without repetition.
f
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d
b
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Example 2.3:
How many 3-letter sequences without repetition and
containing the letter e?
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Use a diagram to display the positions in a
sequence.
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Since the sequence must contain an e, then there
are three choices for which position in the
sequence is e:
e
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e
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Example 2.3 (continued)
How many 3-letter sequences without repetition and
containing the letter e?
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In each diagram, there are 5 choices for which of
the other 5 letters (excluding e) goes in the first
remaining position and 4 choices for which of the
remaining 4 letters goes in the other position.
Thus there are 3*5* 4  60 three-letter sequences
with e.
e
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d
c
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Section 5.1
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Example 2.4:
How many 3-letter sequences with repetition and
containing the letter e?
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As in part 3 when repetition is allowed, there are
three choices for e’s position. For any of these
choices for e’s position, there are 6*6  36
choices for the other two positions, since e and the
other letters can appear more than once. But the
answer of 3*36  108 is not correct. The
Multiplication Principle has been violated because
the outcomes are not distinct.
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Example 2.4 (continued)
How many 3-letter sequences with repetition and
containing the letter e?
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Consider the sequence:
e
c
e
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It was generated two times in our procedure: once when e was
put in the first position followed by c e as one of the 36 choices
for the latter two positions, and a second time when e was put
in the last position with e c in the first two positions.
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We must use an approach that ensures distinct outcomes. Let
us break the problem into disjoint cases based on where the
first e in the sequence occurs.
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Example 2.4 (continued)
How many 3-letter sequences with repetition and
containing the letter e?
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First suppose the first e is in the first position:
a
e
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b
Then there are six choices (including e) for the second and for the third
positions—6 * 6 ways.
Next suppose the first e is in the second position:
d
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Then there are five choices for the first position (cannot be e) and six
choices for the last position—5 * 6 ways.
Finally, let the first (and only) e be in the last position:
c
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f
e
d
e
There are five choices each for the two positions—5 * 5 ways. The correct
answer is thus: (6 * 6)  (5 * 6)  (5 * 5)  91
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Problem for Class to Try:
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How many nonempty different collections can be
formed from five (identical) apples and eight
(identical) oranges?
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Hint: Concentrate on what makes one collection different from another
collection.
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Problem for Class to Try:
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The number of apples and the number of oranges will be
different in different collections.
We can characterize any collections by a pair for integers (a,o),
where a is the number of apples and o is the number of oranges.
There are 6 possible values for a and 9 possible values for o.
Together there are
. Since the problem asked for nonempty collections and
one
of the solutions is (0,0), the desired
6*9
 54
answer is
54  1  53
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Advice:
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When you are stuck and cannot get started with a
problem, try writing down in a systematic fashion of the
possible outcomes you want to enumerate. By doing
this, you should start to see a pattern emerge.
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For help with problem # 45 on the homework, this link
is good:
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http://www.cut-the-knot.com/SimpleGames/FrogsAndToads.shtml
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