Hi.
Skrytý text:We claim that

for all

and prove it by induction.
First of all, let us observe that the sum is nonzero only for

,

and

, i.e. only for such k for which

and

. Such k exists only for

and

, i.e. for

and

. Hence, if we investigate the term

, we find out that
(*)

, where

is the Kronecker delta.
Let us now proceed to the induction and consider

. Then

,
since only

and

are allowed.
Further, let us assume, that the investigated equality is fullfilled for

and we show that it holds for

, as well. Thus, we start with
![kopírovat do textarea $\binom{n}{p}\binom{n}{q} = \left[\binom{n-1}{p}+\binom{n-1}{p-1}\right]\left[\binom{n-1}{q}+\binom{n-1}{q-1}\right] = $](/mathtex/e6/e6c8d8b7932cd593023450baf231482e.gif)

.
Now we apply the inductive step and write



![kopírovat do textarea $=\sum^{n-1}_{k=0}\binom{p}{k}\binom{q}{k}\left[\binom{n+k}{p+q}-{\color{red}\binom{n-1+k}{p+q-1}} \right] + \sum^{n-1}_{k=0}\binom{p-1}{k}\binom{q}{k}{\color{red}\binom{n-1+k}{p+q-1}} +$](/mathtex/ad/ad12c8980dc8284f68ff5f5a724a2bb5.gif)
![kopírovat do textarea $+ \sum^{n-1}_{k=0}\binom{p}{k}\binom{q-1}{k}{\color{red}\binom{n-1+k}{p+q-1}}+ \sum^{n-1}_{k=0}\binom{p-1}{k}\binom{q-1}{k}\left[\binom{n+k}{p+q-1}-{\color{red}\binom{n-1+k}{p+q-1}}\right] =$](/mathtex/26/26718bac2a01939f1d343601795831e7.gif)

![kopírovat do textarea $+ \sum^{n-1}_{k=0} \left[-\binom{p}{k}\binom{q}{k}+\binom{p-1}{k}\binom{q}{k}+\binom{p}{k}\binom{q-1}{k}-\binom{p-1}{k}\binom{q-1}{k}\right]{\color{red}\binom{n-1+k}{p+q-1}}$](/mathtex/8b/8b7ac49771b6b7c0ed876e815a613ed1.gif)
.
The last sum vanishes for

, hence we may consider only

. In addition, if we take into account, that

![kopírovat do textarea $ = -\left[\binom{q}{k}-\binom{q-1}{k}\right]\left[\binom{p}{k}-\binom{p-1}{k}\right] = -\binom{q-1}{k-1}\binom{p-1}{k-1}$](/mathtex/44/4427809d32416c43ad4ff8c08fb8939a.gif)
, then there holds

.
Since for the last sum it holds

,
we obtain

and employing the equality (*) completes the proof.