Greece National Olympiad Problem 3

Lets we assume that every digit from 0 to 9 is a, b, c, d, ... i. Then, we can form an equality a+b+c+d+e+f+g+h+i+j=5 (five is the total of its digits). This is a classic problem. We use the permutation there are 5 objects divide into 10 section that separated by 9 another objects. So it becomes 14!/9!5!. 14C5 is 2002. But there is one solution that isn't accepted when all of its digits are zero. So the answer is 2002-1=2001

But, when you run this code, you get 2001:

   Dim counter As Integer = 0

    For n = 10000 To 99999

        Dim a As String = n.ToString

        Dim a_char() = a.ToCharArray

        If ((a(0) >= a(1)) And (a(1) >= a(2)) And (a(2) >= a(3)) And (a(3) >= a(4))) Then

            counter = counter + 1

        End If

    Next

    Console.WriteLine(counter)

    Console.ReadLine()

$If\quad you\quad follow\quad the\quad pattern\quad of\quad the\quad way\quad these\quad numbers\quad are\quad \\ formed,\quad you\quad can\quad see\quad such\quad a\quad pattern\quad for\quad when\quad the\quad digit\quad \\ { i }\quad is\quad at\quad the\quad 10,000's\quad place:\\ If\quad { k }_{ 1 }(i)\quad represents\quad the\quad number\quad of\quad numbers\quad like\\ \overline { i\quad i\quad i\quad i\quad \_ } ,\quad i.e,\quad the\quad number\quad of\quad numbers\quad with\quad i\quad everywhere\\ but\quad the\quad units\quad place,\quad then\quad \quad { k }_{ 1 }(i)\quad =i+1.\\ Following\quad the\quad weird\quad recursive\quad pattern,\quad { k }_{ 10 }(i)=\sum _{ 0 }^{ i }{ { k }_{ 1 }(r) } \\ =\sum _{ 0 }^{ i }{ (r+1) } =\frac { (i+1)(i+2) }{ 2 } .\\ Here,\quad { k }_{ 10 }(i)\quad represents\quad the\quad number\quad of\quad numbers\quad like\\ \overline { i\quad i\quad i\quad \_ \quad \_ } .\quad Similarly,\quad { k }_{ 100 }(i)=\sum _{ 0 }^{ i }{ \frac { (r+1)(r+2) }{ 2 } } \\ =\sum _{ 0 }^{ i }{ { k }_{ 10 }(r) } =\frac { (i+1)(i+2)(i+3) }{ 6 } .\\ For\quad the\quad last\quad one,\quad { k }_{ 1000 }(i),\quad representing\quad \overline { i\quad \_ \quad \_ \quad \_ \quad \_ } ,\quad we\quad \\ get\quad the\quad number\quad of\quad numbers\quad starting\quad with\quad digit\quad i\quad and\quad \\ subject\quad to\quad given\quad conditions.\\ \\ { k }_{ 1000 }(i)=\sum _{ 0 }^{ i }{ { k }_{ 100 }(r) } =\frac { (i+1)(i+2)(i+3)(i+4) }{ 24 } .\\ Now\quad for\quad the\quad total\quad number\quad of\quad such\quad numbers,\quad \\ N\quad =\quad \sum _{ 1 }^{ 9 }{ { k }_{ 1000 }(r) } =2001.\\ When\quad I\quad entered\quad 2001,\quad it\quad naturally\quad said\quad wrong.\quad I\quad assumed\\ that\quad the\quad question\quad was\quad supposed\quad to\quad include\quad 00000.\quad I\quad entered\quad \\ 2002\quad and\quad it\quad said\quad that\quad it\quad was\quad correct.\\ \\ I\quad rechecked\quad it\quad with\quad a\quad computer\quad program\quad and\quad it\quad said\quad the\quad \\ same\quad thing.\quad I\quad think\quad that\quad the\quad question\quad was\quad meant\quad to\quad read,\\ 'all\quad non-negative\quad integers'.\\ If\quad 2002\quad is\quad the\quad right\quad answer\quad and\quad anyone\quad has\quad obtained\quad it\quad \\ without\quad including\quad 00000,\quad which\quad is\quad not\quad a\quad positive\quad integer,\quad \\ please\quad let\quad me\quad know.$