Is there a short method?

The given series is the coefficient of $x^{15}$ in:

$((1+x)^{17}-1)^{15}$

Now, this can be simplified as:

$x^{15}(17+P(x))^{15}$

Hence, this is simply $17^{15}$ and by elementary modular arithmetic, we can determine the second last digit to be $\boxed{9}$ .

For positive integers $m,n$ , let $X(m,n)$ be the number of surjective (onto) functions from a set with $m$ elements to a set with $n$ elements. Using the Inclusion-Exclusion Principle, we can show that $X(m,n) \; = \; \sum_{r=0}^n (-1)^r {n \choose r} (n-r)^m\,$ so it follows in particular that $\sum_{r=0}^n (-1)^r {n \choose r}(n-r)^m \; = \; \left\{ \begin{array}{lll} 0 & \qquad & 0 \le m < n \\ n! && m = n \end{array} \right.$ An immediate consequence of this is that $\sum_{r=0}^n (-1)^r {n \choose r} r^m \; = \; \left\{ \begin{array}{lll} 0 &\qquad & 0 \le m < n \\ (-1)^n n! && m = n \end{array} \right.$ and hence $\sum_{r=1}^n (-1)^{r-1} {n \choose r}r^m \; =\; \left\{ \begin{array}{lll} 1 & \qquad & m = 0 \\ 0 && 0 < m < n \\ (-1)^{n-1} n! && m = n \end{array} \right.$ (More generally, it is true that $\sum_{r=0}^n (-1)^r {n \choose r} r^m \; = \; (-1)^n n! \left\{ m \atop n \right\}$ where $\left\{ m \atop n \right\}$ is a Stirling number of the second kind, counting the number of partitions of a set with $m$ elements into exactly $n$ subsets.)

Regarded as a function in $r$ , the expression ${Nr \choose n}$ is a polynomial expression in $r$ of degree $n$ with leading term $\frac{N^n}{n!} r^n$ and no constant term. Thus it follows that $\sum_{r=1}^n (-1)^{r-1} {n \choose r} {Nr \choose n} \; = \; (-1)^{n-1}N^n$ for any integers $N,n$ . In our case $n=15$ , $N=17$ , we have that $A \; = \; \sum_{r=1}^{15} (-1)^{r-1} {15 \choose r} {17r \choose 15} \; = \; 17^{15} \;.$ Elementary calculations tell us that $A \,=\, 17^{15} \equiv 93 \pmod{100}$ , and hence that the penultimate digit of $A$ is $\boxed{9}$ .