Question 1135742
<br>NOTE:  To solve this, I got on a train of thought and followed it, but it is likely more optimal solutions exist. <br>

Looking at two-digit numbers A1>A2:

<pre>
A1     A2
9 ---  8,7,...,0  --> 9 numbers in all 
8      8 numbers in all (A2 can be 7,6,...,0)
7      7
...
1      1
     -----
      9+8+...+1 = 9*10/2 = 45 two-digit numbers with A1>A2<br>

I then extended this to three digit numbers A1>A2>A3:
A1     A2
9       8 (which contributes 8 numbers, from the two-digit example)
9       7 (which contributes 7 numbers, again from the two-digit example)

9,A2,A3  has  8+7+6+...+1 = 8*9/2 = 36 numbers  where A1>A2>A3

8,A2,A3  has  7+6+...+1 = 7*8/2 = 28 numbers
7,A2,A3  has  6*7/2 = 21
6,A2,A3  has  5*6/2 = 15
5,A2,A3  has  4*5/2 = 10
4,A2,A3  has  3*4/2 =  6
3,A2,A3  has  2*3/2 =  3
2,A2,A3  has  1*2/2 =  1

36+28+21+15+10+6+3+1 = 120 total three-digit numbers with A1>A2>A3<br>

Now finally four-digit numbers:
9,A2,A3,A4   has 28+21+15+10+6+3+1 = 84 total numbers
(this comes from {three-digit cases for 8,A3,A4} + {three-digit cases for 7,A3,A4} +...+ {three-digit cases for 2,A3,A4})
 
8,A2,A3,A4   has 21+15+10+6+3+1 = 56 total numbers
7,A2,A3,A4   has 15+10+6+3+1 = 35 total numbers
6,A2,A3,A4   has 10+6+3+1 = 20 total numbers
5,A2,A3,A4   has 6+3+1 = 10 total numbers
4,A2,A3,A4   has 3+1 =  4 numbers (4321, 4320, 4310, and 4210)
3,A2,A3,A4   has 1 = 1 number   (3210)

Summing these: 84+56+35+20+10+4+1 = {{{ highlight( 210 )}}}

Check:

Brute-force check by running this Perl one-liner on my MACbook:
perl -e '$c=0; for($i=1000;$i<10000;$i++) { @w=split(//,"$i"); if ($w[0]>$w[1] && $w[1]>$w[2] && $w[2]>$w[3]) { $c++;} } print "count=$c\n";'

210
<br>
And to see all 210 of the numbers with A1>A2>A3>A4:
perl -e '$c=0; for($i=1000;$i<10000;$i++) { @w=split(//,"$i"); if ($w[0]>$w[1] && $w[1]>$w[2] && $w[2]>$w[3]) { $c++; print "$i\n"; } } print "count=$c\n";'