How many eight digit number can be formed by using the digits $1, 2, 3, 4, 5, 6, 7,8$ so that each number has $6$ digits in such place where that digit is less than the next digit?
Example: In number $2314; 2,1$ are two digits such that each of them is less than the next digit.
Regional MO 2016 Dhaka ques 8
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Re: Regional MO 2016 Dhaka ques 8
I think the answer is $7$
Solution: (If anyone check this if it's wrong or not, I'll be happy )
Solution: (If anyone check this if it's wrong or not, I'll be happy )

 Posts: 2
 Joined: Sat Aug 24, 2013 6:46 pm
Re: Regional MO 2016 Dhaka ques 8
isn't the answear (7+6) = 13 ? since, in the question it said how many 'numbers' can be formed. doesn't it mean that there would be different permutations after choosing the 6 digits?? i'm not sure though...
 samiul_samin
 Posts: 999
 Joined: Sat Dec 09, 2017 1:32 pm
Re: Regional MO 2016 Dhaka ques 8
Double post.See here