permuation

usually in permutation without repetition ,the same number wont repeat in the taken value example n= 1,2,1
permutation is n!/(n-r)! here r is 3
so 3!/0!=6 PERMUATION. But what i got in online permutation generator is just 121(no other permutation like 112 ,211...),i mean only one permutation rather than six. Thank you

Denis said:

Can you get someone to help you with your English?
Lots of time is wasted trying to understand what you're posting.

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sorry for inconvenience i caused .This wont happen again.Formula for permutation is n!/(n-r)!.i have took three number of 4,7,4 here we have n=3 and i would like to take r=3 so applying on formula 3!/0!=6 permutation.But i entered the same value in online permutation calculator its showing just only one permutation that is 474.It is not showing the remaining five permuatation.Thank you for correcting

reference online calculator : http://users.telenet.be/vdmoortel/dirk/Maths/permutations.html

bhuvaneshnick said:

have took three number of 4,7,4 here we have n=3 and i would like to take r=3 so applying on formula 3!/0!=6 permutation

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The number string \(\displaystyle 447\) can be rearranged in \(\displaystyle \dfrac{3!}{(2!)}\) ways.

The string \(\displaystyle MISSISSIPPI\) can be rearranged in \(\displaystyle \dfrac{11!}{(2!)(4!)^2}\) ways.
Eleven in all, two P's, four I's & four S's.

pka said:

The number string \(\displaystyle 447\) can be rearranged in \(\displaystyle \dfrac{3!}{(2!)}\) ways.

The string \(\displaystyle MISSISSIPPI\) can be rearranged in \(\displaystyle \dfrac{11!}{(2!)(4!)^2}\) ways.
Eleven in all, two P's, four I's & four S's.

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okay,for now combination we have the formula (combinaiton=permutation/r!). In that case for same 447 ,now for 'r' i am taking take 3( i wish to take 3 position)
so substituting in formula 3/3!= 1/2( chaos )

Here what i have to consider 'r' whether number of position we prefer to take from given number(447) or number of ways that 447 can arranged ?

Thank you

bhuvaneshnick said:

usually in permutation without repetition ,the same number wont repeat in the taken value example n= 1,2,1
permutation is n!/(n-r)! here r is 3
so 3!/0!=6 PERMUATION. But what i got in online permutation generator is just 121(no other permutation like 112 ,211...),i mean only one permutation rather than six. Thank you

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The formula you have is to take r things from n distinct things without replacement where order matters (a permutation), i.e. 121 is different than 112. I'm not sure what online generator you are using but the answer you got appears to be for a combination (order doesn't matter), i.e. 121 is considered the same as 112. If we call the permutation formula
_nP_r = n! / (n-r)!
then the combination formula is
_nC_r = _nP_r / r!

A nice read, IMO, about combinations and permutation with or without replacement including multiple items (not all items are distinct) is given by
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

I understand that you are having trouble understanding how to use the formula and I it is good that you are trying to learn this!

What I do not understand is how you are not seeing how many ways to arrange the three numbers 4, 4, 7.
I see it this way. You can place the 7 in the 1st position (and the 4s in the other two positions) or the 7 in the 2nd position (and the 4s in the other two positions) or the 7 in the 3rd position (and the 4s in the other two positions). In the end there are only 3 ways to permute 4,4,7. This requires no formula at all.
Good luck,
Jomo

Denis said:

true...unless the 4's are of different colors

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Yes, you are correct but my whole point is that since there are all 4s except for one color then you should just see the answer. One should ALWAYS think if there is an obvious answer, like in this case.