考察等式：
\(C\rlap{_{m}}{^{0}}C\rlap{^{r}}{_{n-m}}+C\rlap{_{m}}{^{1}}C\rlap{_{n-m}}{^{r-1}}+…+C\rlap{_{m}}{^{r}}C\rlap{^{0}}{_{n-m}}=C\rlap{_{n}}{^{r}}\)，\((*)\)
其中\(n\)，\(m\)，\(r∈N^{*}\)，\(r\leqslant m < n\)且\(r\leqslant n-m\)．
某同学用概率论方法证明等式\((*)\)如下：设一批产品共有\(n\)件，其中\(m\)件是次品，其余为正品\(.\)现从中随机取出\(r\)件产品，记事件\(A_{k}=\{\)取到的\(r\)件产品中恰有\(k\)件次品\(\}\)，则\(P(A_{k})= \dfrac{C\rlap{_{m}}{^{k}}C\rlap{_{n-m}}{^{r-k}}}{C\rlap{_{n}}{^{r}}}\)，\(k=0\)，\(1\)，\(…\)，\(r.\)显然\(A_{0}\)，\(A_{1}\)，\(…\)，\(A_{r}\)为互斥事件，且\(A_{0}∪A_{1}∪…∪A_{r}=Ω(\)必然事件\()\)，因此\(1=P(Ω)=P(A_{0})+P(A_{1})+…+P(A_{r})= \dfrac{C\rlap{_{m}}{^{0}}C\rlap{^{r}}{_{n-m}}+C\rlap{_{m}}{^{1}}C\rlap{_{n-m}}{^{r-1}}+…+C\rlap{_{m}}{^{r}}C\rlap{^{0}}{_{n-m}}}{C\rlap{_{n}}{^{r}}}\)，所以\(C\rlap{_{m}}{^{0}}C\rlap{^{r}}{_{n-m}}+C\rlap{_{m}}{^{1}}C\rlap{_{n-m}}{^{r-1}}+…+C\rlap{_{m}}{^{r}}C\rlap{^{0}}{_{n-m}}=C\rlap{_{n}}{^{r}}\)，即等式\((*)\)成立．
对此，有的同学认为上述证明是正确的，体现了偶然性与必然性的统一\(.\)但有的同学对上述证明方法的科学性与严谨性提出质疑\(.\)现有以下四个判断：
\(①\)等式\((*)\)成立；\(②\)等式\((*)\)不成立；\(③\)证明正确；\(④\)证明不正确．
试写出所有正确判断的序号：____________．