determinant of pascal matrix- proof Let $U_n$ be the upper triangular Pascal matrix, $L_n$ the lower triangular Pascal matrix of n-th degree, i.e.
$$
 u_{ij} =
  \begin{cases}
    \binom {j-1}{j-i} & \quad i\le j\\
    0  & \quad i>j\\
  \end{cases}
$$
$$
 l_{ij} =
  \begin{cases}
    \binom {i-1}{i-j} & \quad i\ge j\\
    0  & \quad i<j\\
  \end{cases}
$$

The question is how to prove that $P_n=L_n U_n$, where $P_n$ is a matrix, in which the Pascals triangle extends from the upper left corner and is symmetrical with respect to the main diagonal, i.e. 
  $$
p_{ij}=\binom {i+j-2}{j-1}
$$
  Let $[L_n U_n]_{ij}=c_{ij}$, $m=\min(i,j)$, then 
  $$
c_{ij}=\sum_{s =1}^{m}\binom{i-1}{i-s}\binom{j-1}{j-s}=...=(i-1)!(j-1)!\sum_{s =1}^{m}\frac {1}{(i-s)!(j-s)!((s-1)!)^2}
$$
  I am stuck on this step. Could you give me any suggestions how to continue or how to prove the identity $c_{ij}=p_{ij}$ somehow else? Thanks.

 A: @IgorRivin
$$
c_{ij}=\sum_{s =1}^{m}\binom{i-1}{i-s}\binom{j-1}{j-s}=\sum_{s =1}^{m}\binom{i-1}{s-1}\binom{j-1}{j-s}
$$
since $\binom{n}{k}=\binom{n}{n-k}$.

Now to chose $j-1$ elements out of some set of $i+j-2$, first chose $s-1$ out of the first $i-1$, then $ j-s$ out of the other $j-1$. There are $\binom{i-1}{s-1}\binom{j-1}{j-s}$ ways of doing this. Let's mark $M_s$ the set of all such choses. Now if $s_1\ne s_2$ then  $M_{s_1}\cap M_{s_2}=\varnothing$ and if  $s$ goes through all naturals from 1 to $m$ than you get all possible sets $M_s$, i.e. all possible ways of choosing $j-1$ elements out of those $i+j-2$. Therefore 
  $$
\sum_{s =1}^{m}\binom{i-1}{s-1}\binom{j-1}{j-s}=\binom {i+j-2}{j-1}
$$ is that correct?

A: To compute the determinant of $P_n$ without proving this factorization, row reduce $P_n$ to a unipotent upper triangular matrix using Pascal's identity ${i\choose j}+{i\choose j+1} = {i+1\choose j+1}.$ First subtract the $k$-th row from the $k+1$-th row for each $1\leq k\leq n-1$ (starting with $k=n-1$ and moving up), and you will be left with a matrix whose first column is all zeroes except for the diagonal entry. Now subtract the $k$-th row from the $k+1$-th row for each $2\leq k\leq n-1$ in the same way, and continue until you are left with an upper triangular matrix with ones on the diagonal.
A: A combinatorial argument: to chose $j-1$ elements out of $i+j-2,$ you chose $j-s$ from the first $j-1$ elements, and the other $s$ from the second $i-1$ elements, to get
$$
\sum_{s=1}^{j-1} \binom{i-1}{j-s}\binom{j-1}{s} = \binom{i+j-2}{j-1}.$$
Your formula is not quite that, but that may be because you did not do it right..
