The number of four-digit numbers that have distinct digits and are divisible by $99$ We try to find the number of four-digit numbers that have distinct digits and are divisible by $99$.

Let a number be $N = abcd$, then we have $9| N$ and $11|N$.
Thus $9| a+b+c+d$ and $a+c \equiv b+d \mod 11$.
I listed all the four-digit numbers that are divisible by $99$:

From here we can find the numbers with distinct digits. But this method is not elegant.
Is there an elegant method of doing it?

Update: $9| a+b+c+d$ and $a+c \equiv b+d \mod 11$ implies that $9 | a+c$.
Thus $a+c$ is either $9$ or $18$. We can't have $18$ since $a \neq c$.
Hence we have $(a,c) = (1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)$.
 A: Note that:
For a number to be divisible by $99$, the sum of doublets should be divisible by $99$.
Proof:
$abcdef=ab(10^4)+cd(10^2)+ef$ 
and $10^{2n} \mod 99 =1$ 
So, $abcdef \mod 99 = ab + cd + ef$ 
So, $ab+cd+ef$ should be divisible by $99$.
Now let's solve your problem:
Now for a $4$-digit number ($abcd$) to be divisible by $99$ we know that:
$ab+cd=99k$ where $k\in\Bbb Z$ 
Now we can easily see that $k$ must be $1$ as $k=0$ would mean $0000$, invalid and $k=2$ means $9999$, again invalid.
So, we can write $ab+cd=99k$ as $10(a+c) + b+ d = 99$ 
This means that $a+c=9$ and also, $b+d=9$.
Now moving on to cases:
Case $1$: Excluding $0$ (which also means excluding $9$):
We have now $8$ numbers left: $1,2,3,4,5,6,7,8$
Now $a$ can assume any of the $8$ values in $\binom 81$ ways corresponding to which $c$ will automatically assume it's complement.
Since we need all $4$ digits to be distinct so $b$ can now assume any value out of the remaining $6$ digits in $\binom 61$ ways and $d$ will assume it's complement.
So, in total, we get, $\binom 81 \binom 61 = 48$  ways for forming a $4$ digit number in required format.
Case $2$: Including $0$ (and thus $9$):
Now, out of $a$ and $c$, only $c$ can assume $0$ and $a$, $9$. Then $b$ can take any value out of the remaining $8$ digits in $\binom 81$ ways and $d$ will assume it's complement.
Now, out of $b$ and $d$ anyone can assume $0$, in $\binom 21$ ways and then $a$ can take any value from the remaining digits in $\binom 81$ ways, correspondingly $c$ will assume the complement.
So, in total, we get, $\binom 81 + \binom 21\binom 81 = 24$ ways of forming a $4$ digit number in the required format.
Total: (Case $1+$ Case $2$):
Thus, we get $48+24=\bf 72$ as the number of four-digit numbers that have distinct digits and are divisible by $99$.

Checked my result from your image by manual counting too ;) .
