Permutations: Need Help Understanding?

[quantizedzeus]

Permutation...need help...

I don't get the permutation problem about phone numbers...where i can use the same number for several times...or how to put letters in postbox..where i can put all the letters in only one postbox...can anyone explain the general rule and ideas of this kind of permutations...thanks a lot in advance...

 

[JDoolin]

I don't get the permutation problem about phone numbers...where i can use the same number for several times...or how to put letters in postbox..where i can put all the letters in only one postbox...can anyone explain the general rule and ideas of this kind of permutations...thanks a lot in advance...

Well sure. If you have the ability to repeat things its just like counting:

000
001
002
003
004
005

etc. There are 10*10*10 possible three digit numbers.

If you can't repeat, then it's like a tree.

You choose the first item out of 10, then the next item, you've only got 9 choices, then the next item, you've got 8 items. so three digit items with none repeating is:

10 * 9 * 8 = 720 numbers.

That is 10! / (10-3)!

That exclamation point is the factorial sign.

You should also find out about "combinations" as well as "permutations."

 

[LCKurtz]

Think of the "sequential counting principle". If the first step can be done in n ways and the second step can then be done in m ways, then the number of ways the problem can be done is nm ways. The same idea applies to more steps. So if you are filling in a 3 digit code that can have repeats you have 10*10*10 ways. If you can't have repeats you have 10*9*8 ways.

 

[JDoolin]

Excellent Idea LCKurtz.

 

[Ashwin_Kumar]

Permutation is pretty simple actually. Imagine that you have to make a 4 digit number with the numbers 124657 . And you cannot repeat the integers. Then, you simply do 6 per mutate 4, as there are 6 possible integers and you need to select four. The result would be 6*5*4*3, as there are four places.
You can apply this to several problems that also involve arrangement. Take a scenario in which you have some tiles, each 2 by 1 cm thick, and you have to make a 6cm long line. The first scenario would be in which the shorter ends of the tile, the breadth , are arranged side by side. The number of ways to do this would be

$\frac{6!}{6!}=1$

As you have 6 similar tiles and six slots. This is also a form of permutation. Now, let us take a slightly different arrangement. You take 4 breadths and one length( 4 one cm breadths, and a single two cm length will give you 6 cm). The number of ways would be
$\frac{5!}{4!1!}=5$

As you have a total of 5 tiles being used, and 4 breadths and 1 length. So, to find the total number of ways, you simply investigate every single possibility. Another arrangement would be 2 two cm sides and 2 one cm sides. As explained above, the number of ways would be

$\frac{4!}{2!2!}=6$

If you ask me how this is related to the simpler 5P2 permutation, well then it is very similar. 5P2 is basically
$\frac{5!}{3!}$

However, by representing it like how I did, you allow room for more than one factorial in the denominator. Permutation is more complex than combination because in permutation,placement matters, unlike combination, which is just pretty much taking something out of something. Permutation is rather useful as it removes lengthy tables and replaces them with pure math.