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

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

• If the first digit you choose is 0, then you don't actually have a 4-digit number. Oct 7, 2021 at 22:01
• @ConMan That's impossible if the thousands digit is supposed to be the largest digit. Oct 7, 2021 at 22:05
• Right, sorry, I got the > signs the wrong way around. Coffee hasn't kicked in yet. Oct 7, 2021 at 22:06
• 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. Oct 7, 2021 at 22:10
• 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 Oct 8, 2021 at 12:23