6.7 Day 1 Permutations 2011
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January 28, 2011
6.7 Day 1 ‐ Permutations
Use permutations to solve counting problems.
Objectives
6.7 Day 1 Permutations 2011
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January 28, 2011
Warm‐up:
1. Given the word HEAR, how many possible ways are there to arrange the letters?
2. Now take the word HEART. How many possible ways are there to arrange the letters?
6.7 Day 1 Permutations 2011
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January 28, 2011
You were given the word HEAR.
What was your strategy?
How many arrangements did you come up with?
Did you get 24?
What happened when you changed the word to HEART.
What was your strategy?
How many arrangements did you come up with?
Did you get 120?
Is there a pattern??? Is there a trick?? What is it?
6.7 Day 1 Permutations 2011
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There are 4 possible choices for the first letter: H, E, A, and R.
After one of these letters is put in the first space, there are only 3 choices left.
After you have filled spaces 1 and 2, there are only 2 letter choices left.
After you have filled spaces 1, 2, and 3, there is only 1 letter left.
H E A R
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HEAR4 possible choices for the first space
3 possible choices for the second space
2 possible choices for the third space
1 possible choice for the last space
Multiplication Counting Principle: The number of possible outcomes for an event is found by multiplying the number
of choices at each stage of the event.
X X X 1234 = 24
So there are 24 possible arrangements
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Then you added a T. This gave the word HEART. Using the method that we just did, how many different 5letter arrangements can be made from the letters of HEART?
5 x 4 x 3 x 2 x 1 = 120
HEART
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How many different 2letter arrangements can be made from the letters in the word GRATES?
6 x 5 = 30 G RA ET S
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The symbol ! is called factorial and is used in mathematics in a specific way.
You read the symbol 6! as "six factorial". This is what it means:
6! = 6 x 5 x 4 x 3 x 2 x 1Start with the number given to you Mulitply by all the smaller numbers
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PermutationsThe arrangement of any number of items in a
definite order is called a permutation.
The symbol for the number of different arragements when n items are arranged r at a time is Pn rIn the example with HEART, 5 items (letters) are arranged 5 at a time. We used the muliplication counting principle to find the number of possible arrangements.
5 x 4 x 3 x 2 x 1 = 120P5 5 =
5 items arranged 5 at a timeThere are 5 letters, and we are arranging 5 of them
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6 x 5 = 30
We did an example with the word GRATES. We had 6 items (letters) arranged 2 at a time.
P6 2 =
You can also write this using factorials.
P6 2 = 6 x 5 = 6 x 5 x 4 x 3 x 2 x 1 4 x 3 x 2 x 1 = 6!
4!6!
(62)!=
There are 6 letters, and we are arranging 2 of them
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Pn r =n!
(nr)!
In general,
An example,
P10 3 = 10!(103)!
10 x 9 x 8 = 10!7!
=
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January 28, 2011
How many different ways can you arrange a license plate if the first 3 positions are letters and the next 4 positions are numbers 09?
Without repeats,
Letter 1 Letter 2 Letter 3 #1 #2 #3 #4
26 25 24 10 9 8 7
26 x 25 x 24 x 10 x 9 x 8 x 7 = 78,624,000
Letter 1 Letter 2 Letter 3 #1 #2 #3 #4
26 26 26 10 10 10 10
With repeats,
26 x 26 x 26 x 10 x 10 x 10 x 10 = 175,760,000