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Published bySugiarto Hartanto Modified over 5 years ago
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Warm Up β 4/23 - Wednesday There are six colored balls. Two are Red, two are Blue, and two are Green. How many different ways could these balls be arranged in a row?
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Permutation Notation This is the permutation of n objects taken r at a time. For example if we are finding the number of ways 10 runners can finish in 3 places, we would say : π 3
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For the record⦠0!=1
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Permutations with indistinguishable objects
Consider the word MOM with all capital letters. How many ways can I rearrange the letters in the word MOM? If I use the Fundamental Counting theorem I have three letters to choose from and three spots. So I should have 3! Arrangements.
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MOM The problem is that these arrangements include ones where we can distinguish π 1 and π 2 . π 1 π 1 π π 2 π 1 π 1 π 1 π 1 π π 1 π 2 π 1 π 1 π 2 π π 2 π 1 π 1 We get duplicate answers. Our duplicates always result from switching our indistinguishable objects.
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MOM Our answer really is three arrangements. MOM MMO OMM
We get this by dividing by the number of possible switches we can make. In this case 2! 3! 2! =3 arrangments
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Example #2 How many ways can we rearrange the letters in the word EERIE?
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Example #2 - Solution Again we start with : 5 π 5 which gives us 5!.
In this case the letter E is repeated 3 times however, which means we could switch the Eβs 3! times. Thus our answer is: 5! 3! =5β4=20 πππππππππππ‘π .
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Example #3 How many ways can we rearrange the letters in the word ANNA?
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Example #3 - Solution Again we start with : 4 π 4 which gives us 4!.
In this case there are two letters repeated each twice. Which means we will divide by 2! for each letter. Our answer is: 4! 2!2! =3β2=6 ππππππππππ‘π
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Warm Up Revisited For our warm up we have to arrange six objects, so we start with 6!. We will divide by 2! three times because ball can be switched with its twin. There are 6! 2!2!2! = 6β5β4β3β2β1 2β2β2β1β1β1 =3β5β2β3=90
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Permutation Classwork
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