Saying good bye to 2015!

Let us represent the expression $\large\frac{r!}{(r-s)!}$ by $\large P(r,s),$ where $r$ and $s$ are non-negative integer numbers and $r\geq s$ . Then it is easy to see that if $r> s,$ then $\large P(r,s)=\frac{P(r+1, s+1)-P(r, s+1)}{s+1}.\:\:\:\:\:\:\:\:(*)$ Using this formula we obtain $\large \sum_{k=6}^m P(k, 5)=\frac{1}{6}\sum_{k=6}^m(P(k+1, 6)-P(k,6))=\frac{1}{6}(P(m+1,6)-P(6,6)).$ Then $\large N=\sum_{m=6}^{2015} \frac{1}{6}(P(m+1,6)-P(6,6))=\frac{1}{6}\sum_{m=6}^{2015}P(m+1,6)-\frac{1}{6}2010\times P(6,6)$

Applying the formula $(*)$ again we obtain that

$\large N=\frac{1}{6}\sum_{m=6}^{2015}\frac{1}{7}( P(m+2, 7)-P(m+1,7))-\frac{1}{6}2010\times 6!=$ $\large =\frac{1}{42}(P(2017, 7)-P(7, 7)) -2010\times 5!=3200114179550023540440.$ So the answer is : $3+2+0+0+1+1+4+1+7+9+5+5+0+0+2+3+5+4+0+4+4+0=60.$

$\text{Plus}\text{@@}\text{IntegerDigits}\left[\sum _{m=6}^{2015} \sum _{k=6}^m \frac{k!}{(k-5)!}\right]$

$60$

Little value is found from simple routine with only 18 S.F. calculations:

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     N:=0.0;
     FOR m:=6 TO 2015 DO
      FOR k:=6 TO m DO
          N:=N+Cut(1.0*k*(k-1.0)*(k-2.0)*(k-3.0)*(k-4.0));

     WRITELN('Completed.');
     WRITE(N:1:0);
     READLN

3200114179550023540440

Answer: $\boxed{60}$