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February 28th, 2013, 01:26 PM   #1
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Statistics - Permutations (need explanation)

I am trying to do a statistics homework assignment, but I can not seem to grasp the material in the textbook. It is not explained very well, to say the least. I've attached a screenshot of the lesson to this post. I need someone to explain to me (in as simple terms as possible), what this notation really represents. In particular, if someone could tell me what is supposed to go in the spot the author labeled as "..........", it might clear things up for me a bit.

I am not following the pattern of this formula at all.

In particular, this part:

where n! = n(n-1)(n-2)........(3)(2)(1) and 0! = 1 (<<< I have no idea what is meant by this, what it represents, or where these numbers came from)

Also this part:

n! / (n-r)! = n(n-1)(n-2)......(n-r+1)(n-r)....(2)(1) (<<< I can't understand what the "!" is for, or what (n-r+1)(n-r) is, nor what the (2)(1) is for. The explanation provided in my textbook is shown in the attachment. If someone could please explain the top example to me and hopefully show me how it relates to the bottom example, I'd greatly appreciate it.

Thanks.
Attached Images
 permutations.png (61.2 KB, 86 views)

 February 28th, 2013, 02:51 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Statistics - Permutations (need explanation) n! = 1 * 2 * 3 * . . . * n. The dots just mean the pattern is repeated up to n. Example 1: 3! = 1 * 2 * 3 = 6. Example 2: 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720. Example 3: 4! = 1 * 2 * 3 * 4. Example 4: 9! = 1 * 2 * 3 * . . . * 7 * 8 * 9. Consider 6 objects. The number of arrangements for all 6 objects is 6 * 5 * 4 * 3 * 2 * 1 = 720. (6 choices for the first, 5 for the second and so on). Now let's consider how many ways we can arrange 3 of the 6 objects with order taken into account. We have 6 choices for the first, 5 choices for the second and 4 choices for the third. This is equivalent to 6!/3! = 6 * 5 * 4 = 120 ways to order three objects from a set of 6 objects. With n being the number of objects in a set and r being the number of objects we would like to select in any order, the number of permutations, P, is P = n!/(n - r)!. Example: How many permutations of 2 objects from a set of 4 objects? 4!/(4 - 2)! = 4!/2! = (1 * 2 * 3 * 4)/(1 * 2) = 3 * 4 =12. 90!/45! is just a neater way of writing 90 * 89 * 88 * . . . * 48 * 47 * 46. This notation simplifies working with permutations.

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