Collapsing Duplicate Combinations

$ \sum_i \binom{A + i}{i}\binom{B + K - i}{K - i} = \binom{A + B + K + 1}{K}

$ H^n_k = \binom{n + k - 1}{k}

The left side is a duplicate combination that selects i from A+1 and K-i from B+1, and the convolution of this is a duplicate combination that selects K from A+B+2.

$ \sum_i \binom{A + i}{i}\binom{B + K - i}{K - i} = \sum_i H^{A+1}_i H^{B+1}_{K-i} = H^{A+B+2}_K = \binom{A + B + K + 1}{K}

https://gyazo.com/333dc6580cce8f6b5ef4b8cb6abdf495

The problem of adding all patterns under the condition that the lower part of the binomial coefficient is fixed and the upper number has a constant sum.

$ \sum_i \binom{A + i}{i}\binom{B + K - i}{K - i} = \sum_i \binom{A + i}{A}\binom{B + K - i}{B}

Can you make a direct duplicate combination with the lower fixed?

$ \sum_i \binom{A + i}{A}\binom{B + K - i}{B} = \sum_i H_A^{i+1} H_B^{K-i+1} = H_{A+B}^{K+2} = \binom{A + B + K + 1}{A+B}

The math doesn't add up.

This is because the number of frames itself is changing, which is causing double counting.

If we think of adding a bounding ball, the bottom would likely be A+B+1, but then wouldn't we need to increase the top as well? I'm inclined to think that

In fact, this is fine because the boundary does not play the role of only a boundary (the ball can be placed in the same position as the boundary) by being a duplicate combination, but this may be confusing.

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