Can someone explain me something in Permutations??

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h0feer
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#1
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#1
I'm trying to learn Permutations but something is going wrong because i do not understand something in previous tasks that have been done by professor ,



i'm going to upload a photo that i took with my mobile from a task to see what exactly i'm asking for..

Code:
http://speedy.sh/K9tJh/10489126-324965920986474-1240880904-n.jpg

As you can see after the table is done , and after we start working with calculations you can see under 9! permutation are four 2 numbers (2,2,2,2) and after it to 8! under are 2,2,2 at 7! permutation are also 2,2



can someone explain me what they are? what they represent ? how to know where to put four 2s and where three 2's , can someone explain it to me plz plz!


Thanks
Regards
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Notnek
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#2
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#2
(Original post by h0feer)
I'm trying to learn Permutations but something is going wrong because i do not understand something in previous tasks that have been done by professor ,



i'm going to upload a photo that i took with my mobile from a task to see what exactly i'm asking for..

Code:
http://speedy.sh/K9tJh/10489126-324965920986474-1240880904-n.jpg

As you can see after the table is done , and after we start working with calculations you can see under 9! permutation are four 2 numbers (2,2,2,2) and after it to 8! under are 2,2,2 at 7! permutation are also 2,2



can someone explain me what they are? what they represent ? how to know where to put four 2s and where three 2's , can someone explain it to me plz plz!


Thanks
Regards
Can you possibly upload your image instead? There are many free websites where you can upload an image or you could directly upload it to this forum.

I (and probably others) am not comfortable installing a download manager to view your file.
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h0feer
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here's the file bro.



Thanks a lot
Regards
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ghostwalker
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#4
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#4
(Original post by h0feer)
...

The number of permutations on n objects is n!, which I presume what your P(n) is.

If one of those objects is repeated r times say, then you need to divide by r! to get the number of distinct permutations.

E.g. 1,1,1,3,3,4,5,7,8

Here n=9, "1" is repeated 3 times, and "3" is repeated twice.

So, the number of permutations is \dfrac{9!}{3!2!}

2! is of course just 2, and that's where your 2's are coming from.

I'm not entire clear on the method you're using, it seems to be some way of ordering the permutations sequentially.
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h0feer
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#5
Report Thread starter 7 years ago
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(Original post by ghostwalker)
The number of permutations on n objects is n!, which I presume what your P(n) is.

If one of those objects is repeated r times say, then you need to divide by r! to get the number of distinct permutations.

E.g. 1,1,1,3,3,4,5,7,8

Here n=9, "1" is repeated 3 times, and "3" is repeated twice.

So, the number of permutations is \dfrac{9!}{3!2!}

2! is of course just 2, and that's where your 2's are coming from.

I'm not entire clear on the method you're using, it seems to be some way of ordering the permutations sequentially.

ghostwalker,

You are great man, Thanks for your message and your explain

from now on i understand everything about permutations




Thanks a lot
Regards!
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