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    (Original post by DFranklin)
    In the case where you choose the same for both, you don't end up double counting, so dividing the complete total ends up giving the wrong answer.

    E.g. with 3 sweet types A, B, C the options including order are:

    AA, AB, AC
    BA, BB, BC
    CA, CB, CC

    AA, BB and CC only appear once, it's only the cases where we have 2 different letters that we end up double counting (i.e. AB+BA, AC+CA, BC+CB).
    Thats what I thought too which is why I got 210 instead of 200 for 20 possibilities?

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    (Original post by theaverage)
    Thats what I thought too which is why I got 210 instead of 200 for 20 possibilities?

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    Yes.
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    (Original post by RuairiMorrissey)
    Would g(a,b) = s( b-1 , p(a) , p(f(a,m(b)) work for 2015 Q5 part IV?

    The idea being that its one less than f(a,b) which = a + b + 1 as g(a,b) has one less iteration due to x = b-1 instead of x=b
    Did you miss that part (iv) asks for a formula valid for b \leq 0?
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    (Original post by DFranklin)
    Did you miss that part (iv) asks for a formula valid for b \leq 0?
    Ah... yes I did
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    (Original post by DFranklin)
    Did you miss that part (iv) asks for a formula valid for b \leq 0?
    Could I just use my formula but replace b with negative b?

    Zacken


    nvm, i'm wrong end me
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    (Original post by RuairiMorrissey)
    Could I just use my formula but replace b with negative b?
    Your formula also doesn't satisfy the requirement that it uses only the functions p, m and s (because it uses the function f).

    The details of whether you can "just" use your formula but replace b with -ve b depends a bit on exactly how you resolve this. But I'd say "just" isn't going to be a fair description of the process...

    EDIT: TBH, I think the easiest thing is going to be to look at your explanation in (iii), and therefore see how you can craft your own function g using the same ideas.

    2nd EDIT: Just in terms of the "intent" of the question; the general idea is about how to use some really simple operations (adding 1, subtracting 1, choosing between items based on if something is >= 0) to build more complex operations that let you add numbers together. In which case using "-b" in your answer is "cheating", because you've brought in an operation (function) beyond the ones described. (Explicitly, the negation of b is just as much a function as p, m and s). I think it would be harsh to be penalised for this at this level, but I'm sure it's not the desired approach.
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    (Original post by theaverage)
    Thats what I thought too which is why I got 210 instead of 200 for 20 possibilities?

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    Ah i see. You're right.
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    (Original post by DFranklin)
    Your formula also doesn't satisfy the requirement that it uses only the functions p, m and s (because it uses the function f).

    The details of whether you can "just" use your formula but replace b with -ve b depends a bit on exactly how you resolve this. But I'd say "just" isn't going to be a fair description of the process...

    EDIT: TBH, I think the easiest thing is going to be to look at your explanation in (iii), and therefore see how you can craft your own function g using the same ideas.

    2nd EDIT: Just in terms of the "intent" of the question; the general idea is about how to use some really simple operations (adding 1, subtracting 1, choosing between items based on if something is >= 0) to build more complex operations that let you add numbers together. In which case using "-b" in your answer is "cheating", because you've brought in an operation (function) beyond the ones described. (Explicitly, the negation of b is just as much a function as p, m and s). I think it would be harsh to be penalised for this at this level, but I'm sure it's not the desired approach.
    I ended up going for g(a,b) = s( p(b) , p [g ( a , p(b) ) ], p(a) )

    Not sure if this would work, the mark scheme has a similar, but slightly different answer.
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    (Original post by RuairiMorrissey)
    I ended up going for g(a,b) = s( p(b) , p [g ( a , p(b) ) ], p(a) )

    Not sure if this would work, the mark scheme has a similar, but slightly different answer.
    You should try it and see (say with a=5, b=-2, this time).

    But I'm pretty sure it isn't going to work, you're going to need to use the 'm' function somewhere in your answer I'm sure.
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    2013 Q2 ii
    I got it when I did this question but now I am just looking over the answers/questions and can't remember the reasoning behind it.. can someone explain
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    How do you:

    Work out the questions for digit sum: e.g. How many positive integers n such that n<500 have a digit sum of 9?

    Is there a formula to work it out as it gets really boring to work it out for big numbers.

    Someone suggested to me its 500/9 and then round it to the nearest integer but im not sure if it works.
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    for 2009 Q2 iv why do they only add up to n-1 instead of n terms?
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    (Original post by Mystery.)
    for 2009 Q2 iv why do they only add up to n-1 instead of n terms?
    Hmm?
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    (Original post by RuairiMorrissey)
    Hmm?
    Am I being stupid again?
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    (Original post by Mystery.)
    Am I being stupid again?
    No no, i'm just not sure what you're asking, it's me being stupid haha
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    (Original post by RuairiMorrissey)
    No no, i'm just not sure what you're asking, it's me being stupid haha
    Well from what I can see they have left the 1 added the 2, 4, ...2n
    by the formula n/2(first term + last term) but instead of having n it is upto n-1?
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    (Original post by Mystery.)
    Well from what I can see they have left the 1 added the 2, 4, ...2n
    by the formula n/2(first term + last term) but instead of having n it is upto n-1?
    If you're referring to the number of terms, because they're not including the 1 there are n-1 terms when going up to Yn.
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    (Original post by RuairiMorrissey)
    If you're referring to the number of terms, because they're not including the 1 there are n-1 terms when going up to Yn.
    Yeah but they aren't taking the one into account anyway no? Because they used 4 as the first term
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    (Original post by Mystery.)
    Yeah but they aren't taking the one into account anyway no? Because they used 4 as the first term
    Exactly, so there's just n-1 terms in the sequence

    I feel as though i'm misunderstanding your question, sorry!
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    Can anyone help with specimen B question 4 i)
 
 
 
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