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Working With Permutations:

A.  Permutations are often confused with combinations.

1.  A permutation can be thought of as changing the alignment of a group.

2.  In other words, if we had 3 people A, B, and C, one possible permutation can be ABC, while another can be CAB. 

3.  Permutations come in many forms. 

a.   Some forms are obvious permutations:

6P3 or P(6,3)

b.  Other forms are not so obvious and are written as word problems which is somewhat difficult to recognize.

Ex [1]  How many ways can 3 people be seated 3 at a time in 4 chairs?

Ex [2]  How many ways can 2 people sit in 5 chairs in a row?

1)  One way to know if we are dealing with permutations or combinations, is to answer one question:  Does the order matter?

2)  If the answer is no, then we will be using permutations, not combinations.

Ex [1] How many ways can 3 people be seated 3 at a time in 4 chairs?

a.  We have 3 people: A, B, and C.

b.  Does the order matter?  Can I have ABC, CBA, ACB, CAB, BAC, and BCA and count them as 6 or do they all count as 1.  In this example it would count as 6, so the order does not matter.

In Ex [2] we can use the same reasoning to see that order will not matter in that case either.

B.  How to calculate a permutation:

1.  This method uses factorials.

2.  P(n,r) =    n!


Ex [1]  5P2 = __________.

  1. 5!/(5-2)! = 5*4*3*2*1/3*2*1  

  2. 5*4*3*2*1/3*2*1 = 5*4 = 20.

  3. The answer is 20.

Ex [2]  How many ways can 4 people sit in 6 chairs in a row?  __________.

  1. Using the same reasoning from above, we can see that order does not matter, so it is indeed a permutation problem.

  2. Notice, we will be using P(6,4).

  3. 6!/(6-4)! = 6*5*4*3*2*1/2*1  

  4. 6*5*4*3*2*1/2*1 = 720/2 = 360.

  5. The answer is 360.

*Note:  Sometimes it is easier to cancel some numbers out before multiplying as in Ex [1] step b.  Other times it might be easier to compute the numerator first, then the denominator, then divide as in Ex [2] step d.


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