basically im curious as to the methods for questions like "how many combinations" etc
i know if i have a question were i have n variables with a positions i can use a^n
binary for example, 8  bits with 2 options, 0 and 1  there are 2^8 combinations available
but what if i can't repeat combinations?
such as questions like, i have 10p 20p 50p etc etc
how many combinations of change can i make? (you obviously can't repeat any combinations)
or the question which made me think of this type of maths was "If I had a cube and six colours and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had n colours instead of 6?"
any reading i should do? what topic would this fall under?

Chaoslord
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 09112008 19:43

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 09112008 20:04
Take three girls and three seats.
How many girls can possibly sit on the first seat?
When one girl is seated, how many girls are left over who can possibly sit on the second seat? 
WhiteAndNerdy1729
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 09112008 20:08
It depends if order matters in the context you're working in. If you had a group of people and wanted to pick of them in any order, the number of combinations is (read n choose r) which is a shorthand way of saying . Some people write this as .
If the order matters, we call the options 'permutations' and denote the number of them as which is equivalent to . Hope this helps! 
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 09112008 20:09
Oops, sorry Rocious I gave the game away a bit there! Chaoslord, see if you can use Rocious' post to derive either or both of the formulae I gave!

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 09112008 20:11
(Original post by Rocious)
Take three girls and three seats.
How many girls can possibly sit on the first seat?
When one girl is seated, how many girls are left over who can possibly sit on the second seat?
3 girls can sit on seat on
given one is sat, there are 7 ways for the other girls to sit
0 = no girl
1 = girl one
2 = girl two
00
01
10
02
20
12
21
(3^3  2) since there is 22 and 11 that cant work
7 x 3 (since there are 3 girls who can sit in seat one) = 21 
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 09112008 20:14
(Original post by WhiteAndNerdy1729)
It depends if order matters in the context you're working in. If you had a group of people and wanted to pick of them in any order, the number of combinations is (read n choose r) which is a shorthand way of saying . Some people write this as .
If the order matters, we call the options 'permutations' and denote the number of them as which is equivalent to . Hope this helps!
i've seen that before when using polynomials
how can i decide if order matters? 
Horizontal 8
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 09112008 20:17
The topic i suppose would be combinatorics?
In my pure maths textbook there is a chapter on this type of stuff (things like combinations and permutations). Its quite interesting stuff for example this is a question on circular arrangements:
How many different ways can you arrange 4 people around a circular table? Well if u fix the position of the first person then you have 3 spots to fill in with the 3 other people. For the first spot you have a choice of 3 people, for the second you have a choice of 2 and for the third a choice of one. Hence the answer is 3 factorial i.e. 6.
In general the answer would be (n1)! for n people around a circular table.
However, I have also seen this topic in my stats texbook and I think its currently in some stats modules. (The reason its in my pure one aswell is because its an old one )
Hope this helps 
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 09112008 20:20
yea i normally do the same thing, fix on point and then find it from there
but for a cube you can't just fix one point and then multiply by 6 if i have 6 colours since this could repeat some cubes
would i just not multiply by 6? 
Horizontal 8
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 09112008 20:23
(Original post by Chaoslord)
i've seen that before when using polynomials
how can i decide if order matters?
EDIT: For the cube one i think it would be 6!. Assuming you can distinguish between the sides otherwise its just 1
The way I see it there are six sides take one (i.e. any) side first, and you have a choice of six colours. There are 5 sides left, pick any of these and you have a choice of 5 colours.. etc
6 x 5 x 4 x 3 x 2 x 1 = 6!
For n colours you have : n x (n1) x (n2) x (n3) x (n4) x (n5) ways of painting the cube
(Once again I think)
This is linked to the formula nPr and this idea can be used to derive it.Last edited by Horizontal 8; 09112008 at 20:42. 
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 09112008 20:41
(Original post by WhiteAndNerdy1729)
It depends if order matters in the context you're working in. If you had a group of people and wanted to pick of them in any order, the number of combinations is (read n choose r) which is a shorthand way of saying . Some people write this as .
If the order matters, we call the options 'permutations' and denote the number of them as which is equivalent to . Hope this helps! 
abstraction98
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 09112008 20:43
Seconded (And r should be subtext in both cases just as the n is supertext

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 09112008 20:47
(Original post by abstraction98)
Seconded (And r should be subtext in both cases just as the n is supertext 
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 09112008 20:55
(Original post by Horizontal 8)
sorry My latex skills are lazy, infact I admit it I copied it from his post which in fact makes me lazy as opposed to my skills ... 
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 09112008 21:07
haha i assumed xD 3!/3! = 1 =P
also the thing with the cubes can be confusing, i normaly would say 6! but, if you say fix one point, and chose the next and so on then you could get a rotation of a cube which has already been accounted for
think about it like this, pick a colour x and chose the other sides, if all the 4 of the sides change could and only 2 remain constant they cube is not a new cube it is a rotation on a used one. 
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 09112008 21:42
(Original post by Chaoslord)
haha i assumed xD 3!/3! = 1 =P
also the thing with the cubes can be confusing, i normaly would say 6! but, if you say fix one point, and chose the next and so on then you could get a rotation of a cube which has already been accounted for
think about it like this, pick a colour x and chose the other sides, if all the 4 of the sides change could and only 2 remain constant they cube is not a new cube it is a rotation on a used one.
However my working was made assuming you could for example (although unecessary), call the sides:
S_1, S_2 , ..., S_6 in order to distinguish between them. 
WhiteAndNerdy1729
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 09112008 21:58
Ooops I typed the post in a hurry and missed that  sorry Chaoslord! And yes I missed the subscript too. rep!

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 09112008 22:00
(Original post by Horizontal 8)
Im not sure i fully understand what you mean but from what i understood the main problem is the fact that unless you can distinguish each side i.e. assign a particular side a particular colour, there is only one combination since all other combinations of 6 colours are just rotations and you cant tell the difference between each side so they all look the same.
However my working was made assuming you could for example (although unecessary), call the sides:
S_1, S_2 , ..., S_6 in order to distinguish between them.
but there are ways of compiling a dice such that it doesn't that being put the 6 on a side touching the 1 and theres a few ways =P 
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 09112008 22:37
(Original post by Chaoslord)
its not one, take a dice for example it has a format such that all opposite sides sum to 7
but there are ways of compiling a dice such that it doesn't that being put the 6 on a side touching the 1 and theres a few ways =P
Infact in this case i think you can fix one of the sides and so the number of ways would be (n1)! i.e. 5! but not so sure so please don't take much of what i say seriously i only skimmed over a chapter on this stuff 
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 09112008 22:41
(Original post by Horizontal 8)
Yes, its not one because this is a case where one can distinguish between the sides since they're numbered
Infact in this case i think you can fix one of the sides and so the number of ways would be (n1)! i.e. 5! but not so sure so please don't take much of what i say seriously i only skimmed over a chapter on this stuff 
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 09112008 22:53
(Original post by Chaoslord)
haha no problem, im gonna do some research tomorrow =]
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