PERMUTATIONS AND COMBINATIONS

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PERMUTATIONSANDCOMBINATIONS
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PERMUTATIONS AND COMBINATIONS ARE METHODS
TO SOLVE CERTAIN TYPES OF WORD PROBLEMS.
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BOTH PERMUTATIONS AND COMBINATIONS USE A
COUNTING METHOD CALLED FACTORIAL.
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A FACTORIAL is a counting method that uses
consecutive whole numbers as factors. The
factorial symbol is ! Examples 5! 5x4x3x2x1
120
7! 7x6x5x4x3x2x1
5040
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First, well do some permutation
problems. Permutations are arrangements.
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Lets do a permutation problem. How many
different arrangements are there for 3 books on a
shelf? Books A,B, and C can be arranged in these
ways ABC ACB BAC BCA CAB CBA Six
arrangements or 3! 3x2x1 6
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In a permutation, the order of the books is
important. Each different permutation is a
different arrangement. The arrangement ABC is
different from the arrangement CBA, even though
they are the same 3 books.
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You try this one 1. How many ways can 4 books
be arranged on a shelf?
4! or 4x3x2x1 or 24 arrangements
Here are the 24 different arrangements ABCD
ABDC ACBD ACDB ADBC ADCB BACD
BADC BCAD BCDA BDAC
BDCA CABD CADB CBAD CBDA CDAB
CDBA DABC DACB DBAC DBCA
DCAB DCBA
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Now were going to do 3 books on a shelf again,
but this time were going to choose them from a
group of 8 books.

• Were going to have a lot more possibilities this
  time, because there are many groups of 3 books to
  be chosen from the total 8, and there are several
  different arrangements for each group of 3.

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If we were looking for different arrangements for
all 8 books, then we would do 8!

• But we only want the different arrangements for
  groups of 3 out of 8, so well do a partial
  factorial,
• 8x7x6
• 336

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Try these 1. Five books are chosen from a group
of ten, and put on a bookshelf. How many
possible arrangements are there?
10x9x8x7x6 or 30240
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2. Choose 4 books from a group of 7 and arrange
them on a shelf. How many different arrangements
are there?
7x6x5x4 or 840
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Now, well do some combination problems. Combinat
ions are selections.
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There are some problems where the order of the
items is NOT important.
These are called combinations.
You are just making selections, not making
different arrangements.
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Example A committee of 3 students must be
selected from a group of 5 people. How many
possible different committees could be formed?
Lets call the 5 people A,B,C,D,and E.
Suppose the selected committee consists of
students E, C, and A. If you re-arrange the
names to C, A, and E, its still the same group
of people. This is why the order is not
important.
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Because were not going to use all the possible
combinations of ECA, like EAC, CAE, CEA, ACE, and
AEC, there will be a lot fewer committees.
Therefore instead of using only 5x4x3, to get the
fewer committees, we must divide.
(Always divide by the factorial of the number of
digits on top of the fraction.)
Answer 10 committees
5x4x3 3x2x1
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Now, you try.
1. How many possible committees of 2 people can
be selected from a group of 8?
8x7 2x1
or 28 possible committees
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2. How many committees of 4 students could be
formed from a group of 12 people?
12x11x10x9 4x3x2x1
or 495 possible committees
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CIRCULAR PERMUTATIONS
When items are in a circular format, to find the
number of different arrangements, divide
n! / n
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Six students are sitting around a circular table
in the cafeteria. How many different seating
arrangements are there?
6! ? 6 120
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THE END