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    The theory says that if you have x objects containing a repeats of one element and b repeats of another, Np(without repetition)=x!/(a!b!).

    If you have x objects and repetitions are allowed, Np(with repetition)=xx, correct?

    Combining these, if we have x objects containing a repeats of one element and b repeats of another, and repetitions are allowed, Np(with repetition)=xx/(a!b!). Is this right?
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    (Original post by Big-Daddy)
    The theory says that if you have x objects containing a repeats of one element and b repeats of another, Np(without repetition)=x!/(a!b!).

    If you have x objects and repetitions are allowed, Np(with repetition)=xx, correct?
    I presume you mean x different objects, in which case yes, that's the number of arrangements.

    Combining these, if we have x objects containing a repeats of one element and b repeats of another, and repetitions are allowed, Np(with repetition)=xx/(a!b!). Is this right?
    Nope.

    The formula will be similar to x^x, but not quite. Have another go. Consider what each "x" represents there.
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    (Original post by ghostwalker)
    I presume you mean x different objects, in which case yes, that's the number of arrangements.



    Nope.

    The formula will be similar to x^x, but not quite. Have another go. Consider what each "x" represents there.
    Since it's you, let's just work with my old notation.

    Hmm how about x^(x-r-s)? I feel like I'm shooting in the dark a bit here. The index represents how many spaces we have to fill; the base represents number of elements we have. Maybe (n-a-b)^(x-r-s) multiplied by ar multiplied by bs?
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    (Original post by Big-Daddy)
    Since it's you, let's just work with my old notation.

    Hmm how about x^(x-r-s)? I feel like I'm shooting in the dark a bit here. The index represents how many spaces we have to fill; the base represents number of elements we have.
    Your explanation is correct, but you changed the wrong x in the formula - :p:

    Maybe (n-a-b)^(x-r-s) multiplied by ar multiplied by bs?
    Going back to your full quesion then. As I said on the other thread, this is more complex. I'll have to think about it.
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    This may take a while, since we're preselecting some of the items, before looking at the permutations with repetition. And we can get overlap between the permutations from different selections.
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    (Original post by ghostwalker)
    This may take a while, since we're preselecting some of the items, before looking at the permutations with repetition. And we can get overlap between the permutations from different selections.
    Huh. So C(n-a-b,x-r-s)*(x-r-s)^x isn't enough?
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    (Original post by Big-Daddy)
    Huh. So C(n-a-b,x-r-s)*(x-r-s)^x isn't enough?
    Considering the original question of this thread, you want (x-r-s+2)^x since there are x-r-s singular ones, plus the two types that occur r,s times - forgot about the 2 previously.

    Considering the question from your other thread. It's not a question of being enough. It's too many. Since different choices from the original set, can lead to the same permutation. Every way I've thought about it, gets messier and messier. So, I don't intend to spend any more time on it. I think the way forward might be to use inclusion/exclusion on the number of types selected.

    It's not something you'll need for A-level.
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    (Original post by ghostwalker)
    Considering the original question of this thread, you want (x-r-s+2)^x since there are x-r-s singular ones, plus the two types that occur r,s times - forgot about the 2 previously.

    Considering the question from your other thread. It's not a question of being enough. It's too many. Since different choices from the original set, can lead to the same permutation. Every way I've thought about it, gets messier and messier. So, I don't intend to spend any more time on it. I think the way forward might be to use inclusion/exclusion on the number of types selected.

    It's not something you'll need for A-level.
    I see, thanks for all your help.
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    (Original post by ghostwalker)
    Considering the question from your other thread. It's not a question of being enough. It's too many. Since different choices from the original set, can lead to the same permutation. Every way I've thought about it, gets messier and messier. So, I don't intend to spend any more time on it. I think the way forward might be to use inclusion/exclusion on the number of types selected.

    It's not something you'll need for A-level.
    How would P(n-a-b,x-r-s)*(x!/(r!s!(x-r-s)!) be? Does that do it? (Had a hint from a friend)
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    (Original post by Big-Daddy)
    How would P(n-a-b,x-r-s)*(x!/(r!s!(x-r-s)!) be? Does that do it? (Had a hint from a friend)
    Try it on some small values and see.

    I don't see any value in you just posting different answers and then asking me to confirm/deny. You can post your working and justification for each step, and I/someone can check if it's reasonable. Though, if your friend knows the answer then it would be better for you to deal with them.
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    (Original post by ghostwalker)
    Try it on some small values and see.

    I don't see any value in you just posting different answers and then asking me to confirm/deny. You can post your working and justification for each step, and I/someone can check if it's reasonable. Though, if your friend knows the answer then it would be better for you to deal with them.
    You're right, this was unreasonable. I will discuss it with my friend (and insist that he takes me through it rather than just handing it to me.)
 
 
 
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