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Math Help - Factorials

  1. #1
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    Factorials

    Hi,

    Does anyone know whether it is possible to simplify a factorial as follows:

    {n\choose i}{n-m\choose i-m}

    = (n-m)! / ((i-m)!(n-m-i-m)!)
    =(n-m)!/((i-m)!(n-i)!)
    =(n-m)!/(i-m+n-i)!
    =(n-m)!/(n-m)!
    =1
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Number Cruncher 20 View Post
    Does anyone know whether it is possible to simplify a factorial as follows:

    {n\choose i}{n-m\choose i-m}

    = (n-m)! / ((i-m)!(n-m-i-m)!)
    No... your forgot \tbinom{n}{i}=\tfrac{n!}{i!(n-i)!}. We have

    \begin{aligned}<br />
{n\choose i}{n-m\choose i-m}&=\frac{n!}{i!(n-i)!}\cdot \frac{(n-m)!}{(i-m)![n-m-(i-m)]!}\\&=\frac{n!}{i!(n-i)!}\cdot \frac{(n-m)!}{(i-m)!(n-i)!}\\&= \frac{(n-m)!n!}{i!(i-m)!(n-i)!^2}\end{aligned}

    and it seems this can't be further simplified. If you were asked to simplify \tbinom{n}{m}\tbinom{n-m}{i-m} then we could derive the equality \tbinom{n}{m}\tbinom{n-m}{i-m}=\tbinom{n}{i}\tbinom{i}{m} but that's not what you're asked to do, is it?
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  3. #3
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    no, what ive been asked is to derive a closed formed expression for

    (a) s = for n= 0,1,2,....

    (b) t = for n=0,1,2, .. and m = 0,1,2,...n

    ive done the first part and showed that its equal to 1 when n = 0 and 0 when n not equal to 0. for the second part ive been given a hint that

    for .

    and to use this and part a to find a closed form expression for (b). i get the part a and the hint but i just dont see how to go from there to solving part b.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Number Cruncher 20 View Post
    and to use this and part a to find a closed form expression for (b). i get the part a and the hint but i just dont see how to go from there to solving part b.
    We have

     t=\sum_{i=m}^n(-1)^i\binom{n}{i}\binom{i}{m}=\sum_{i=m}^n(-1)^i\binom{n}{m}\binom{n-m}{i-m}<br />

    since we are told that \tbinom{n}{i}\tbinom{i}{m}=\tbinom{n}{m}\tbinom{n-m}{i-m}. As \tbinom{n}{m} doesn't depend on i we can factor this term to get

     t=\binom{n}{m}\sum_{i=m}^n(-1)^i\binom{n-m}{i-m}.

    Using the result of question (a), can you find a closed formed expression for \sum_{i=m}^n(-1)^i\binom{n-m}{i-m} and thus for t ?

    Hint (highlight to read) : * change the index of summation to j = i-m *
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  5. #5
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    I havent done summations for a while. Have i changed the variables properly so that we get:


     <br />
\binom{n}{m}\sum_{j=o}^n(-1)^(j+m)\binom{n-m}{j}<br />
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Number Cruncher 20 View Post
    I havent done summations for a while. Have i changed the variables properly so that we get:


     <br />
\binom{n}{m}\sum_{j=o}^n(-1)^(j+m)\binom{n-m}{j}<br />
    That's almost it, there is only a little mistake : for i=n,\,j=n-m so

    t=\binom{n}{m}\sum_{j=0}^{{\color{blue}n-m}}(-1)^{j+m}\binom{n-m}{j}<br />
.

    Factoring (-1)^m we get

    t=(-1)^m\binom{n}{m}\sum_{j=0}^{n-m}(-1)^j\binom{n-m}{j}<br />

    and now you can use the answer to the first question to conclude.
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  7. #7
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    Thank you for your help its really appreciated.
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  8. #8
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    One last thing, but should the sum read:

    (-1)^(j+2m) or is it right as (-1)^(j+m)
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Number Cruncher 20 View Post
    One last thing, but should the sum read:

    (-1)^(j+2m) or is it right as (-1)^(j+m)
    If you're talking about the following sum
    Quote Originally Posted by flyingsquirrel View Post
    t=\binom{n}{m}\sum_{j=0}^{{\color{blue}n-m}}(-1)^{j+m}\binom{n-m}{j}<br />
    then it is right as (-1)^{j+m} since this term comes from (-1)^{i} where we've substituted i=j+m. Why do you think it should be (-1)^{j+2m} ?
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  10. #10
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    sorry my bad i forgot the restrictions on the range of n and m so i thought it had to be (-1)^{2m} for it to be 1.

    Thanks again
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  11. #11
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    what i meant is that you get
     <br /> <br />
(-1)^{m} \binom{n}{m} (0)^{n-m}<br />

    and when n=m

    if m and n are odd then t = -1 and when m and n are even then t is 1

    is that the right answer.
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