Showing two series are equal to each other I would like to show that these two series are equal to each other:
$\sum_j {m+n-j\choose j}$=$\sum_j {m-j\choose j}$$\sum_j {n-j\choose j}$ + $\sum_j {m-j-1\choose j}$$\sum_j {n-j-1\choose j}$.
By Applying Pascal's rule to the first pair of sums on the R.H.S, where Pascal's Rule is:
${n\choose k}={n-1\choose k}+{n-1\choose k-1}$
and setting ${n\choose k}={m-j\choose j}$
I get ${m-j\choose j}={m-j-1\choose j}+{m-j-1\choose j-1}$ and ${n-j\choose j}={n-j-1\choose j}+{n-j-1\choose j-1}$
This gives R.H.S=$(\sum_j {m-j-1\choose j}+{m-j-1\choose j-1}) $$(\sum_j {n-j-1\choose j}+{n-j-1\choose j-1}$) + $\sum_j {m-j-1\choose j}$$\sum_j {n-j-1\choose j}$.
By expanding the brackets I find:
R.H.S=$$2\sum_j {m-j-1\choose j}\sum_j {n-j-1\choose j}+\sum_j {m-j-1\choose j-1}\sum_j {n-j-1\choose j}+\sum_j {m-j-1\choose j}\sum_j {n-j-1\choose j-1}+\sum_j {m-j-1\choose j-1}\sum_j {n-j-1\choose j-1}$$
Is this correct so far? And are there any hints you can give me to get the R.H.S equal to the L.H.S?
 A: I have found it useful to interpret binomial coefficient $\binom{a}{b}$ as the number of right/up paths in a rectangular grid to the point $[a-b, b]$. In this case, the sum $\sum_j \binom{n-j}{j}$ represents the number of paths ending at some point $[n-2j, j]$.
The trick is to realize, what on the other hand you have to do if you want to avoid such points of form. You have to go to a point $[n-2j-1, j]$, then you have to go upwards to $[n-2j-1, j+1]$ and then you can continue.
With these observations, the identity can be just realized. You can also use this idea to think about partial results, it may help to see whether you are approaching the identity.
Why is the identity clear:
LHS = the number of paths to a point $[n+m-2i, i]$.
Let's divide these paths on those passing through a point of the form $[n-2j, j]$ and those what don't. If you want to pass such a point, you have to go to one: $\sum_j\binom{n-j}{j}$ possibilities. Now, there is another key point: the pattern of points $[n+m-2i, i]$ looks the same from every point of form $[n-2j, j]$. You finalize your path by going $m-2k$ to the right and $k$ upwards, where $k=i-j$: $\sum_i\binom{n-k}{k}$ possibilities. Together, there are $\sum_i\binom{n-j}{j}\sum_i\binom{n-k}{k}$ paths passing through a point of form $[n-2j, j]$.
The non-passing paths can be computed similarly. First you have to get to a point of form $[n-2j-1, j]$, $\sum_j\binom{n-j-1}{j}$ possibilities. Then go upwards and finally by $[m-2k-1, k]$: $\sum_k\binom{n-k-1}{k}$ possibilities.
By the way, I used the same technique for a similar question.
