That’s a beautiful picture and exactly what I was talking about, Omphaloskeptic. Wendell, I think you missed the ‘copied’ lines that I added, which change the sum.

Just to make sure I have this right:

The sum of the actual chunk taken out of the triangle is equal (almost) to a “southwest” diagonal of length k+1 starting at the end of row n-1. The ‘almost’ is needed because the first value (1) is not included in the sum. But that will get subtracted out later, so for now I’ll go with this as the sum of that box.

In terms of the coefficients, we have sum(i=0->k) B(n-1+i,n-1), which using the identity for a diagonal in this direction yields B(n+k,n). The diagram shows that this is the value that the parallelogram ‘points to’ at its bottom (that is, the next value vertically down from the lowest corner).

```
/\
/ 1\
/ \
/ 1 1\
/ \
| 1 2 1\
\ \
1 \ 3 3 1\
\ \
1 4 \ 6 4 1|
\ /
1 5 10 \10 5/ 1
\ /
1 6 15 20 \15/ 6 1
\/
1 7 21 35 35 21 7 1
+--+
1 8 28 56 70 |56| 28 8 1
+--+
The sum of the boxed values is equal to the marked coefficient (56), less 1.
```

The two copied legs are equal to the number just below and to the right of the box (for the ‘success’ leg), and just below and to the left for the other one. These are 35 and 21 in the 5 x 3 example. But the sum of these two is just the same single binary coefficient for the box’s sum B(n+k,n).

Therefore the total values in my special ‘near-parallelogram’ are

Total = 2*B(n+k,n) - 1

with the extra 1 being subtracted out.

To put this in terms of the total ‘successes’, note that s = B(n+k-1,n).

Dividing gives us:

## B(n+k,n)

B(n+k-1,n)

## (n+k)!/((n+k-n)!n!)

(n+k-1)!/((n+k-1-n)!n!

## (n+k)!(k-1)!

(n+k-1)!k!

## (n+k)

k

Thus B(n+k,n) = (n+k)/k * B(n+k-1,n) and the formula makes sense.

Thanks for the help.