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    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?
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    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?
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    It depends if order matters in the context you're working in. If you had a group of n people and wanted to pick r of them in any order, the number of combinations is ^nCr_ (read n choose r) which is a shorthand way of saying \frac{n!}{r!(n-r)!}. Some people write this as ${n \choose r}$.

    If the order matters, we call the options 'permutations' and denote the number of them as ^nPr_ which is equivalent to \frac{n!}{r!}. Hope this helps!
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    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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    (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?
    is this case you can't use 3^3

    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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    (Original post by WhiteAndNerdy1729)
    It depends if order matters in the context you're working in. If you had a group of n people and wanted to pick r of them in any order, the number of combinations is ^nCr_ (read n choose r) which is a shorthand way of saying \frac{n!}{r!(n-r)!}. Some people write this as ${n \choose r}$.

    If the order matters, we call the options 'permutations' and denote the number of them as ^nPr_ which is equivalent to \frac{n!}{r!}. Hope this helps!

    i've seen that before when using polynomials

    how can i decide if order matters?
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    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 (n-1)! 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 :p:)

    Hope this helps
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    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?
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    (Original post by Chaoslord)
    i've seen that before when using polynomials

    how can i decide if order matters?
    For permutations order matters however for combinations order doesn't matter. It's usually easy to know what to use from the question , for example if the question is on how many numbers enjoy a certain property (e.g. the sum of the individual digits add up to 39. Yes i know its a step question!) then the order of the digits matters (clearly!) since different orders give different numbers.

    EDIT: For the cube one i think it would be 6!. Assuming you can distinguish between the sides otherwise its just 1 :p:
    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 (n-1) x (n-2) x (n-3) x (n-4) x (n-5) 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.
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    (Original post by WhiteAndNerdy1729)
    It depends if order matters in the context you're working in. If you had a group of n people and wanted to pick r of them in any order, the number of combinations is ^nCr_ (read n choose r) which is a shorthand way of saying \frac{n!}{r!(n-r)!}. Some people write this as ${n \choose r}$.

    If the order matters, we call the options 'permutations' and denote the number of them as ^nPr_ which is equivalent to \frac{n!}{r!}. Hope this helps!
    By the way your ^nPr_ formula is wrong its actually \frac{n!}{(n-r)!}
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    (Original post by Horizontal 8)
    By the way your ^nPr_ formula is wrong its actually \frac{n!}{(n-r)!}
    Seconded (And r should be subtext in both cases just as the n is supertext
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    (Original post by abstraction98)
    Seconded (And r should be subtext in both cases just as the n is supertext
    sorry :p: 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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    (Original post by Horizontal 8)
    sorry :p: 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 ...
    No worries. I don't even know how to do latex. Need to put that on my todo list
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    (Original post by Horizontal 8)
    By the way your ^nPr_ formula is wrong its actually \frac{n!}{(n-r)!}
    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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    (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.
    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.
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    (Original post by Horizontal 8)
    By the way your ^nPr_ formula is wrong its actually \frac{n!}{(n-r)!}
    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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    (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.
    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
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    (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
    Yes, its not one because this is a case where one can distinguish between the sides since they're numbered :p:
    Infact in this case i think you can fix one of the sides and so the number of ways would be (n-1)! 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 :o:
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    (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 :p:
    Infact in this case i think you can fix one of the sides and so the number of ways would be (n-1)! 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 :o:
    haha no problem, im gonna do some research tomorrow =]
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    (Original post by Chaoslord)
    haha no problem, im gonna do some research tomorrow =]
    Its a really interesting topic.. And often something you read on it makes questions soo much easier (including STEP questions). Have fun
 
 
 
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