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Math Help - Combinatory Counting help

  1. #1
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    Combinatory Counting help

    A protein may contain several cysteins, which my pair together to form disulfide bonds as shown in the figure. If there is an even number n of cysteins, n/2 disulfide bonds can form. How many different disulfide pairing arrangements are possible?

    For clarity, see attached for the question with the figure.

    So I think the number of arrangements, N, would be equal to:

    N=(n/2)!

    Is this correct, and if not, could someone explain it to me? Thanks,

    Kim
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  2. #2
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    Quote Originally Posted by Kim Nu View Post
    A protein may contain several cysteins, which my pair together to form disulfide bonds as shown in the figure. If there is an even number n of cysteins, n/2 disulfide bonds can form. How many different disulfide pairing arrangements are possible?

    For clarity, see attached for the question with the figure.

    So I think the number of arrangements, N, would be equal to:

    N=(n/2)!

    Is this correct, and if not, could someone explain it to me? Thanks,

    Kim
    Hi Kim,

    I am not sure I understand the problem correctly, but the statement seems to say there are n/2 possible bonds, each of which may or may not form-- so there are 2 possibilities for each possible bond.

    If this is so, then the total possible number of arrangements is 2^{n/2}.
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  3. #3
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    The problem is not very clearly defined, and I have a different interpretation from awkward's.

    Based on the drawing, it seems to me that a configuration is given by a choice of "non-crossing pairings".

    In other words, a configuration is a 2p-uplet (i_1,\ldots,i_p,j_1,\ldots,j_p) such that 0\leq p\leq n/2 and:

    1\leq i_1< \cdots < i_{p-1}< i_p < j_p < j_{p-1} <\cdots < j_1 \leq n

    (The bonds are between the cysteins at positions i_q and j_q for q=1,\ldots,p, cf. the attached drawing).

    As a consequence, the number of configurations is the number of subsets of \{1,\ldots,n\} with an even number of elements (say, 2p).

    Thus the number of configurations is:

    N=\sum_{0\leq 2p\leq n}{n\choose 2p}.

    In fact, there are as many subsets of \{1,\ldots,n\} with an even or an odd number of elements. This is a consequence of the following equality (using the binomial formula):

    \sum_{0\leq 2p\leq n}{n\choose 2p}-\sum_{0\leq 2p+1\leq n}{n\choose 2p+1} =\sum_{0\leq k\leq n}(-1)^k {n\choose k} = (1-1)^n =0.

    We deduce that the number of even-sized subsets of \{1,\ldots,n\} is half the number of subsets of \{1,\ldots,n\}.

    As a conclusion, N=\frac{1}{2}2^n = 2^{n-1}.
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    Last edited by Laurent; February 14th 2009 at 04:04 AM.
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  4. #4
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    Laurent,

    With your formula, N=2^(n-1), for 4 molecules the total number of different pairing options would be 2^3 = 8. I don't see how 8 different pairing options could be possible, could you explain it to me? Thanks,

    Kim
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  5. #5
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    Quote Originally Posted by Kim Nu View Post
    Laurent,

    With your formula, N=2^(n-1), for 4 molecules the total number of different pairing options would be 2^3 = 8. I don't see how 8 different pairing options could be possible, could you explain it to me? Thanks,

    Kim
    My interpretation allows one bond between adjacent molecules; this may provide the configurations you're missing (there are 6 configs with 1 bond, 1 config with 2 bonds, and 1 with 0 bond). I admit this kind of pairings is a bit weird, relative to the chemistry problem... After all, perhaps awkward got it right, I don't know.
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