Question 1150894
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If 2 is in the hundreds place, then there is only 1 three-digit number that is a descending number and it is 210.


If 3 is in the hundreds place, then there are three such descending numbers
310
320
321


If 4 is in the hundreds place, then we have six descending numbers
410
420
421
430
431
432



If 5 is in the hundreds place, then we have ten descending numbers
510
520
521
530
531
532
540
541
542
543


The counts so far are: 1, 3, 6, 10


The jump from 1 to 3 is +2
The jump from 3 to 6 is +3
The jump from 6 to 10 is +4


This pattern continues because we have 15 descending numbers if 6 is the hundreds digit
610
620
621
630
631
632
640
641
642
643
650
651
652
653
654


And there are 15+6 = 21 descending numbers with 7 as the hundreds digit, 21+7 = 28 descending numbers with 8 as the hundreds digit, and finally 28+8 = 36 descending numbers with 9 as the hundreds digit.


The counts are
1,3,6,10,15,21,28,36
Add them up
1+3+6+10+15+21+28+36 = 120


Answer: <font color=red size=4>120</font>


Below is a pastebin link of all 120 three-digit descending numbers
<a href = "https://pastebin.com/WqFYtfc4">https://pastebin.com/WqFYtfc4</a>
This is to avoid adding further clutter to this current page.
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