Each number $\overline{a_k \ldots a_1a_0}$ maps to $\overline{(9-a_k) \ldots (9-a_1)(9-a_0)}$. Thus, number of $x$ so $S(x)=9(n-2)$ and $x<10^n$ equal to number of $x$ so $S(x)=18$ and $x<10^n$. This is also equal to number of ordered $n$-tuples of non-negative integers $(x_1, \ldots, x_n)$ so $x_1+\ldots +x_n=18$ and $x_i \le 9$. If we neglect the condition $x_i \le 9$ (for all $i$) then number of such tuples is $\tbinom{n+17}{18}$. We then take out the tuples that has one $x_i>9$. Indeed, if there is one $x_i=k>9$ then $\sum_{j \ne i} x_j=18-k<9$. This gives us a result of $\tbinom{n-2+18-k}{n-2}$ ordered $n-1$-tuples and there are $n$ ways to put $x_i=k$ in. Hence, there are $n\cdot \tbinom{n-k+16}{n-2}$ $n$-tuples so that there exists one $x_i=k>9$. Thus, number of such $n$-tuples so $\sum_{i=1}^n x_i=18$ and $0 \le x_i \le 9$ or number of such $x$ is \begin{align*} \binom{n+17}{18}-n\sum_{k=10}^{18}\binom{n-k+16}{n-2} & =\binom{n+17}{18}-n \sum_{k=0}^8 \binom{n-2+k}{n-2}, \\ & = \binom{n+17}{18}-n \cdot \binom{n+7}{8}. \end{align*}Number of $x$ so $S(x)=9(n-1)$ and $x<10^n$ equal to number of $x$ so $S(x)=9$ and $x<10^n$. This is equal to number of $n$-tuples $(x_1, \ldots, x_n)$ so $\sum_{i=1}^n x_i=9$, which is $\tbinom{n+8}{9}$. Thus, it suffices to prove that $$\binom{n+8}{9}< \binom{n+17}{18}-n \cdot \binom{n+7}{8}.$$This is true for $n>3$.

shinichiman wrote: Each number $\overline{a_k \ldots a_1a_0}$ maps to $\overline{(9-a_k) \ldots (9-a_1)(9-a_0)}$. Thus, number of $x$ so $S(x)=9(n-2)$ and $x<10^n$ equal to number of $x$ so $S(x)=18$ and $x<10^n$. This is also equal to number of ordered $n$-tuples of non-negative integers $(x_1, \ldots, x_n)$ so $x_1+\ldots +x_n=18$ and $x_i \le 9$. If we neglect the condition $x_i \le 9$ (for all $i$) then number of such tuples is $\tbinom{n+17}{18}$. We then take out the tuples that has one $x_i>9$. Indeed, if there is one $x_i=k>9$ then $\sum_{j \ne i} x_j=18-k<9$. This gives us a result of $\tbinom{n-2+18-k}{n-2}$ ordered $n-1$-tuples and there are $n$ ways to put $x_i=k$ in. Hence, there are $n\cdot \tbinom{n-k+16}{n-2}$ $n$-tuples so that there exists one $x_i=k>9$. Thus, number of such $n$-tuples so $\sum_{i=1}^n x_i=18$ and $0 \le x_i \le 9$ or number of such $x$ is \begin{align*} \binom{n+17}{18}-n\sum_{k=10}^{18}\binom{n-k+16}{n-2} & =\binom{n+17}{18}-n \sum_{k=0}^8 \binom{n-2+k}{n-2}, \\ & = \binom{n+17}{18}-n \cdot \binom{n+7}{8}. \end{align*}Number of $x$ so $S(x)=9(n-1)$ and $x<10^n$ equal to number of $x$ so $S(x)=9$ and $x<10^n$. This is equal to number of $n$-tuples $(x_1, \ldots, x_n)$ so $\sum_{i=1}^n x_i=9$, which is $\tbinom{n+8}{9}$. Thus, it suffices to prove that $$\binom{n+8}{9}< \binom{n+17}{18}-n \cdot \binom{n+7}{8}.$$This is true for $n>3$. Hey man, you said that $$\sum_{k=10}^{18}\binom{n-k+16}{n-2}=\sum_{k=0}^8 \binom{n-2+k}{n-2}$$. But in my calculations, I found $$\sum_{k=10}^{18}\binom{n-k+16}{n-2}=\sum_{k=0}^8 \binom{n+6-k}{n-2}$$. How did you find this result? Thanks

NiltonCesar wrote: Hey man, you said that $$\sum_{k=10}^{18}\binom{n-k+16}{n-2}=\sum_{k=0}^8 \binom{n-2+k}{n-2}$$. But in my calculations, I found $$\sum_{k=10}^{18}\binom{n-k+16}{n-2}=\sum_{k=0}^8 \binom{n+6-k}{n-2}$$. How did you find this result? Thanks They are both correct. Note that I used $+k$ but yours is $-k$.

shinichiman wrote: NiltonCesar wrote: Hey man, you said that $$\sum_{k=10}^{18}\binom{n-k+16}{n-2}=\sum_{k=0}^8 \binom{n-2+k}{n-2}$$. But in my calculations, I found $$\sum_{k=10}^{18}\binom{n-k+16}{n-2}=\sum_{k=0}^8 \binom{n+6-k}{n-2}$$. How did you find this result? Thanks They are both correct. Note that I used $+k$ but yours is $-k$. Great, man. Now i'm understand. Thanks again