Decreasing sequence numbers with first digit $9$ 
Find the sum of all positive integers whose digits (in base ten) form a strictly decreasing sequence with first digit $9$.

The method I thought of for solving this was very computational and it depended on a lot of casework. Is there a nicer way to solve this question?
Note that there are $\displaystyle\sum_{n=0}^{9} \binom{9}{n} = 2^9$ such numbers.

For each digit $a$ from $0$ to $9$, let us count how many numbers there are in our sum with a digit of $a$ in the $10^n$ place.  First, suppose $a<9$.  A number with a digit of $a$ in the $10^n$ place has a subset of $\{0,1,\dots,a-1\}$ for its last $n$ digits, so there are $\binom{a}{n}$ choices for the last $n$ digits.  The preceding digits (omitting the initial $9$) can form any subset of $\{8,7,\dots,a+1\}$, so there are $2^{8-a}$ choices for the preceding digits.  So in total, all of the digits of $a$ in our numbers contribute $\sum_n 2^{8-a}\binom{a}{n}\cdot a10^n$ to the sum.  By the binomial theorem, this is equal to $$2^{8-a}a(10+1)^a=2^8a\left(\frac{11}{2}\right)^a.$$
For $a=9$, we have no choice of preceding digits, so we just have $\binom{9}{n}$ choices.  So the sum of the $9$ digits contributes $$\sum_n \binom{9}{n}9\cdot 10^n=9\cdot 11^9.$$
So in total, the sum is $$9\cdot 11^9+2^8\sum_{a=0}^8 a\left(\frac{11}{2}\right)^a.$$
Let us write $x=\frac{11}{2}$ and simplify the sum in the second term:
$$\sum_{a=0}^8 ax^a=\sum_{b=1}^8\sum_{c=b}^8x^c=\sum_{b=1}^8\frac{x^9-x^b}{x-1}=\frac{8x^9-\frac{x^9-x}{x-1}}{x-1}.$$
Putting it all together, the original sum is
$$9\cdot 11^9+2^8\cdot\frac{8x^9-\frac{x^9-x}{x-1}}{x-1}$$
for $x=\frac{11}{2}$.  According to Wolfram Alpha, this evaluates to $23259261861$.  That is, assuming I haven't made some algebra mistake somewhere.
