Weighted sum of binomial coefficients choose three

Compute $S=\sum_{m=3}^{14}(15-m)\binom{m}{3}=12\binom33+11\binom43+\cdots+2\binom{13}{3}+1\binom{14}{3}.$

$4368$. Split the weight $15-m$ into that many copies and apply the hockey-stick identity twice: $\sum_{j=3}^{14}\sum_{m=3}^{j}\binom m3=\sum_{j=3}^{14}\binom{j+1}4=\binom{16}{5}=4368$.