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4.3 Permutations with Some Identical Elements
There are three different type of permutations:
a) Permutations of all elements; P(n,n) = n!
Example: What are the total number of arrangements for the following six balls if all
balls must be used?
b) Permutations involving some elements from the population;
Example: What are the total number of arrangements for the following six balls if only
three balls must be used?
These two types were demonstrated in the previous lesson.
c) Permutations involving some identical elements;
Example: What are the total number of arrangements for the following: 3 blue balls, 2 red balls, and 1 green ball?
Explanation:
The total number of permutations for the arrangement of the six balls is 6!, but these are not distinct permutations. The distinction between the permutations within three blue balls and also between the two red balls is not possible, resulting in repetition which includes non-distinct permutations. In order to eliminate this repetition, the total number of permutations are divided by the permutations of the repeated elements.
Therefore, resulting in:
Example 1: How many distinct permutations can be made from the name AJAY?
The total number of permutations for the name AJAY are 4!.
These are not distinct permutations, the permutations of the identical letters (elements), “A”, do not affect the total number of distinct permutations possible.
Therefore, the number of distinct permutations for AJAY does not equal 4!
To eliminate the non-distinct permutations, the total number of permutations are divided by the permutations of the repeated identical elements.
Therefore, the total number of distinct permutations for AJAY are:
Example 2: How many 7-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6?
This questions involves cases:
Case 1: First digit is 1 and last digit is 6.
Case 2: First digit is 2 and last digit is 6.
Case 3: First digit is 2 and last digit is 2.
Case 4: First digit is 1 and last digit is 2.
Now use the Additive Counting Principle (ACP): (4.1 Organized Counting)
Therefore, there are a total 210 distinct permutations.
The process for finding the number of permutations can be summarized in this flow chart:
some of the permutations for a Rubik cube that lead to the solution of the cube.
References:
Mathematics of Data Management , Grade 12, (MDM4U). McGraw-Hill Ryerson
http://www.youtube.com
http://www.gliffy.com
http://www.imeem.com
Mathematics of Data Management , Grade 12, (MDM4U). McGraw-Hill Ryerson
http://www.youtube.com
http://www.gliffy.com
http://www.imeem.com
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Keyword tags:
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| Anonymous | xfcgf | 0 | Apr 10 2012, 2:34 PM EDT by Anonymous | ||
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| Anonymous | PERMUTATIONS | 2 | Feb 14 2012, 11:47 AM EST by Anonymous | ||
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Thread started: Feb 3 2012, 8:51 AM EST
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How many 2 letter words can be formed from letters of the word " SPEECH" ?
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| Anonymous | Example 2 done differently: pls. tell if this correct | 0 | Dec 25 2011, 1:33 PM EST by Anonymous | ||
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Thread started: Dec 25 2011, 1:33 PM EST
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I was using the concept that you explained where we divide by the common permutations
Case 1: First digit is 1 and last digit is 6 1st and last positions are taken by 1 and 6. Now we have positions 2 to 6 left to be filled by 2,2,3,5,5 = total unique permutations/Common Permutation cases = 5!/(2! x 2!) Comments: there are two 2s and two 5s Case 2: First digit is 1 and last digit is 2 1st and last positions are taken by 1 and one of 2s. Now we have positions 2 to 6 left to be filled by 2,3,5,5,6 = total unique permutations/Common Permutation cases = 5!/(2!) Comments: there are two 5s. One of the 2 is tied up in position 7, so only one 2 was left for middle. 5 repeats twice Case 3: First digit is 2 and last digit is 6 1st and last positions are taken by one of 2 and 6. Now we have positions 2 to 6 left to be filled by 1,2,3,5,5 = total unique permutations/Common Permutation cases = 5!/(2!) Comments: there are two 5s. One of the 2 is tied up in position 1, so only one 2 was left for middle. 5 repeats twice Case 4: First digit is 2 and last digit is 2 1st and last positions are taken by 2 and 2. Now we have positions 2 to 6 left to be filled by 1,3,5,5,6 = total unique permutations/Common Permutation cases = 5!/(2!) Comments: there are two 5s. One of the 2 is tied up in positions 1 and 7. 5 repeats twice This will result in 60 + 120 + 120 + 120 = 420 cases. Please let me know if this understanding correct. |
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