Regional MO 2016 Dhaka ques 8

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Md. Rifat uddin
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Joined: Thu Dec 15, 2016 1:32 pm

Regional MO 2016 Dhaka ques 8

Unread post by Md. Rifat uddin » Thu Dec 15, 2016 7:40 pm

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.

super boy
Posts: 5
Joined: Tue Dec 20, 2016 1:11 pm

Re: Regional MO 2016 Dhaka ques 8

Unread post by super boy » Sun Dec 25, 2016 12:44 pm

I think the answer is $7$

Solution: (If anyone check this if it's wrong or not, I'll be happy :) )
Notice when the symbol $ "<" $ is inserted in the appropriate place which is between the eight digits, there are $7$ place to put that symbol inside the eight digits. Also we've given a condition that "each number has $6$ digits in such place where that digit is less than the next digit", so I've to put $6$ symbols in that $7$ places. I can do that $\binom{7}{6} = 7$ such ways. And that's the answer I think. :/

Nodee Haque
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Joined: Sat Aug 24, 2013 6:46 pm

Re: Regional MO 2016 Dhaka ques 8

Unread post by Nodee Haque » Sun Jan 01, 2017 2:55 am

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...

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samiul_samin
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Joined: Sat Dec 09, 2017 1:32 pm

Re: Regional MO 2016 Dhaka ques 8

Unread post by samiul_samin » Mon Feb 18, 2019 1:54 pm

Double post.See here

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