Hi,
Does anyone know whether it is possible to simplify a factorial as follows:
$\displaystyle {n\choose i}{n-m\choose i-m}$
= (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
Hello,
No... your forgot $\displaystyle \tbinom{n}{i}=\tfrac{n!}{i!(n-i)!}$. We have
$\displaystyle \begin{aligned}
{n\choose i}{n-m\choose i-m}&=\frac{n!}{i!(n-i)!}\cdot \frac{(n-m)!}{(i-m)![n-m-(i-m)]!}\\&=\frac{n!}{i!(n-i)!}\cdot \frac{(n-m)!}{(i-m)!(n-i)!}\\&= \frac{(n-m)!n!}{i!(i-m)!(n-i)!^2}\end{aligned}$
and it seems this can't be further simplified. If you were asked to simplify $\displaystyle \tbinom{n}{m}\tbinom{n-m}{i-m}$ then we could derive the equality $\displaystyle \tbinom{n}{m}\tbinom{n-m}{i-m}=\tbinom{n}{i}\tbinom{i}{m}$ but that's not what you're asked to do, is it?
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.
We have
$\displaystyle 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}
$
since we are told that $\displaystyle \tbinom{n}{i}\tbinom{i}{m}=\tbinom{n}{m}\tbinom{n-m}{i-m}$. As $\displaystyle \tbinom{n}{m}$ doesn't depend on $\displaystyle i$ we can factor this term to get
$\displaystyle t=\binom{n}{m}\sum_{i=m}^n(-1)^i\binom{n-m}{i-m}$.
Using the result of question (a), can you find a closed formed expression for $\displaystyle \sum_{i=m}^n(-1)^i\binom{n-m}{i-m}$ and thus for $\displaystyle t$ ?
Hint (highlight to read) : * change the index of summation to j = i-m *
That's almost it, there is only a little mistake : for $\displaystyle i=n,\,j=n-m$ so
$\displaystyle t=\binom{n}{m}\sum_{j=0}^{{\color{blue}n-m}}(-1)^{j+m}\binom{n-m}{j}
$.
Factoring $\displaystyle (-1)^m$ we get
$\displaystyle t=(-1)^m\binom{n}{m}\sum_{j=0}^{n-m}(-1)^j\binom{n-m}{j}
$
and now you can use the answer to the first question to conclude.
If you're talking about the following sum
then it is right as $\displaystyle (-1)^{j+m}$ since this term comes from $\displaystyle (-1)^{i}$ where we've substituted $\displaystyle i=j+m$. Why do you think it should be $\displaystyle (-1)^{j+2m}$ ?