More fun in 2016, Part 14

i am not good in combo so i use a calculus approach. consider the summation $\sum_{k=0}^{2016} (-1)^{2016-k}(x+1)^k \binom{2016}{k}=x^{2016}$ we get this by the binomial theorem. we d.w.r.x and multiply both sides with (x+1) to get $\sum_{k=0}^{2016} (-1)^{2016-k}(x+1)^k k \binom{2016}{k}=2016x^{2015}(x+1)$ we repeat this 2015 times! we dont have to do that. we just easily notice that the polynomial formed after 2015 differentiation will not have any constant term or at x=0. that will be zero. basically $\sum_{k=0}^{2016} (-1)^{2016-k} k^{2015}\binom{2016}{k}=0$ or $\sum_{k=2}^{2016} (-1)^{2016-k} k^{2015}\binom{2016}{k}=\boxed{2016}$

We can interpret this combinatorially. Note that if we shift the index so that it starts from $k=0$ , by the principle of inclusion and exclusion, this is equivalent to arranging 2015 balls into a row of 2016 boxes such that each box has a positive number of balls. This is obviously zero. Thus, the desired expression is just ${2016 \choose 1}=2016$ .

Distinct Objects into Distinct Bins