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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?
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.
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|Anonymous||How would this be tackled using your line of reasoning||2||Nov 22 2012, 1:03 PM EST by Anonymous|
Thread started: Aug 21 2012, 3:16 PM EDT Watch
A group of 5 friends—Archie, Betty, Jerry, Moose, and Veronica—arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?
|Anonymous||math||0||Aug 7 2012, 10:40 PM EDT by Anonymous|
|Anonymous||xfcgf||0||Apr 10 2012, 2:34 PM EDT by Anonymous|
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