1. ## 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

2. Hello,
Originally Posted by Number Cruncher 20
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}
{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?

3. 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.

4. Originally Posted by Number Cruncher 20
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}
$

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 *

5. I havent done summations for a while. Have i changed the variables properly so that we get:

$
\binom{n}{m}\sum_{j=o}^n(-1)^(j+m)\binom{n-m}{j}
$

6. Originally Posted by Number Cruncher 20
I havent done summations for a while. Have i changed the variables properly so that we get:

$
\binom{n}{m}\sum_{j=o}^n(-1)^(j+m)\binom{n-m}{j}
$
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}
$
.

Factoring $(-1)^m$ we get

$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.

7. Thank you for your help its really appreciated.

8. One last thing, but should the sum read:

(-1)^(j+2m) or is it right as (-1)^(j+m)

9. Originally Posted by Number Cruncher 20
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
Originally Posted by flyingsquirrel
$t=\binom{n}{m}\sum_{j=0}^{{\color{blue}n-m}}(-1)^{j+m}\binom{n-m}{j}
$
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 $(-1)^{j+2m}$ ?

10. 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

11. what i meant is that you get
$

(-1)^{m} \binom{n}{m} (0)^{n-m}
$

and when n=m

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