How many four digit numbers are such that reading from right to left , each digit is greater than the digit at previous position?

My approach:

We want a 4 digit number of type abcd where a>b>c>d

According to me answer should be 10c4 , as any 4 numbers chosen will have only 1 arrangement in which a>b>c>d, but the answer given in the book is 9c4, please help me identify my mistake

  • $\begingroup$ If the first digit you choose is 0, then you don't actually have a 4-digit number. $\endgroup$
    – ConMan
    Oct 7, 2021 at 22:01
  • $\begingroup$ @ConMan That's impossible if the thousands digit is supposed to be the largest digit. $\endgroup$
    – Arthur
    Oct 7, 2021 at 22:05
  • $\begingroup$ Right, sorry, I got the > signs the wrong way around. Coffee hasn't kicked in yet. $\endgroup$
    – ConMan
    Oct 7, 2021 at 22:06
  • 4
    $\begingroup$ Looks to me like your solution is elegant and correct. I encourage you to include the problem statement in the body of the Question, not just in the title. $\endgroup$
    – hardmath
    Oct 7, 2021 at 22:10
  • 2
    $\begingroup$ the books answer is true for $a<b<c<d$ ,but if the question $d<c<b<a$ , then for $abcd$ your answer is correct $\endgroup$ Oct 8, 2021 at 12:23


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