Consider the following equality of symmetric polynomials of degree $n$:
$\sum\limits_{i=1}^{n-1}c_ix^iy^{n-i}+\sum\limits_{i=1}^{n-1}c_i(x+y)^iz^{n-i}=\sum\limits_{i=1}^{n-1}c_iy^iz^{n-i}+\sum\limits_{i=1}^{n-1}c_ix^{i}(y+z)^{n-i}$
(We set $c_0=c_n=0$ and $c_i=c_{n-i}$.)
Now how can one see that (after comparing coefficients on both sides) we get for the coefficients of $x^{\alpha}y^{\beta}z^{n-\alpha-\beta}, \alpha >0, \alpha + \beta <n$,
$c_{\alpha + \beta} \binom{\alpha + \beta}{\beta}=c_{\alpha}\binom{n-\alpha}{\beta}$?
Assuming all exponents are positive we have general terms \begin{eqnarray*} LHS = c_i \binom{i}{j}x^{i-j} y^jz^{n-i} \end{eqnarray*} and \begin{eqnarray*} RHS = c_I x^{I} \binom{n-I}{n-J} y^{n-I-J}z^{J}. \end{eqnarray*} This gives \begin{eqnarray*} i-j&=&I&=&\alpha \\ j&=&n-I-J&=&\beta \\ n-i&=&J&=&n-\alpha-\beta. \end{eqnarray*} Now solve these ($j=\beta,i=\alpha+\beta,n-I=n-\alpha,n-J=n-\alpha-\beta$) and we have \begin{eqnarray*} c_{\alpha+\beta} \binom{\alpha+\beta}{\beta}=c_{\alpha}\binom{n-\alpha}{n-\alpha-\beta} \end{eqnarray*} the second binomial coefficient is the same as is stated in the question by symmetry of binomial coefficients.