3/26/03 tucker, applied combinatorics section 5.11 5.1 general counting methods for arrangements and...

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3/26/03 Tucker, Applied Combinato rics Section 5.1 1 5.1 General Counting Methods for Arrangements and Selections Tucker, Applied Combinatorics, Sec. 5.1, Patti Bodkin and Tamsen Hunter

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Page 1: 3/26/03 Tucker, Applied Combinatorics Section 5.11 5.1 General Counting Methods for Arrangements and Selections Tucker, Applied Combinatorics, Sec. 5.1,

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5.1 General Counting Methods for Arrangements and Selections

Tucker, Applied Combinatorics, Sec. 5.1, Patti Bodkin and Tamsen Hunter

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Principals

The Addition PrincipalIf there are different objects in the first set, objects in the second set, . . . , and the

objects inthe mth set, and if the different sets are

disjoint,then then number of the ways to select an

objectfrom one of the m sets is .

1r 2rmr

1 2 ... mr r r

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Principals

The Multiplication PrincipalSuppose a procedure can be broken into msuccessive (ordered) stages, with outcomesin the first stage, outcomes in the second stage,. . . , and outcomes in the mth stage. If thenumber of outcomes at each stage is independent

ofthe choices in previous stages and if the compositeoutcomes are all distinct, then the total procedurehas different composite outcomes.

1r2r

mr

1 2... mr r r

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Example 1There are:

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

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

How many ways are there to form a three-letter sequence using the letters a, b, c, d, e, f:

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?

With repetition, we have six choices for each letter in the sequence. So by the multiplication principle there are three-letter sequences without repetition.

2166*6*6

a a b

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Example 2.2:How many 3-letter sequences without repetition?

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 three-letter sequences without repetition.

1204*5*6

f d b

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Example 2.3:How many 3-letter sequences without repetition and containing the letter e?

Use a diagram to display the positions in a sequence.

Since the sequence must contain an e, then there are three choices for which position in the sequence is e:

eee

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Example 2.3 (continued)

How many 3-letter sequences without repetition and containing the letter e?

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 three-letter sequences with e.

604*5*3

e d c

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Example 2.4:How many 3-letter sequences with repetition and

containing the letter e?

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 choices for the other two positions, since e and the other letters can appear more than once. But the answer of is not correct. The Multiplication Principle has been violated because the outcomes are not distinct.

366*6

10836*3

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Example 2.4 (continued)

How many 3-letter sequences with repetition and

containing the letter e? Consider the sequence:

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.

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.

e c e

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Example 2.4 (continued)

How many 3-letter sequences with repetition and

containing the letter e?

First suppose the first e is in the first position:

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:

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:

There are five choices each for the two positions—5 * 5 ways. The correct answer is thus: 91)5*5()6*5()6*6(

e a b

ed f

e c d

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Problem for Class to Try:

How many nonempty different collections can be formed from five (identical) apples and eight (identical) oranges?

Hint: Concentrate on what makes one collection different from another collection.

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Problem for Class to Try:

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

non-empty collections and one of the solutions is (0,0), the desired answer is 53154

549*6

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Advice:

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.

For help with problem # 45 on the homework, this link is good:

http://www.cut-the-knot.com/SimpleGames/FrogsAndToads.shtml