Dhaka Higher Secondary 2010/9
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can 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.
Dhaka Higher Secondary 2010/9
Out of the digits $1, 2, 3, 4, 5$, three are chosen to form numbers so that their digits are either in increasing or decreasing order. What is the total number of numbers formed? If one number is chosen from each category so that the sum of those two numbers is maximum, what is that sum?
 leonardo shawon
 Posts: 169
 Joined: Sat Jan 01, 2011 4:59 pm
 Location: Dhaka
Re: Dhaka Higher Secondary 2010/9
decreasing order 543 > 432 > 321
increasing order  123 < 234 < 345
the sum is 888
increasing order  123 < 234 < 345
the sum is 888
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: Dhaka Higher Secondary 2010/9
The second part is correct. However, you need to show some logic. Actually we are trying maximize the number. So apparently the largest number $543$ works.
Hint for the first part:
Hint for the first part:
"Inspiration is needed in geometry, just as much as in poetry."  Aleksandr Pushkin
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 leonardo shawon
 Posts: 169
 Joined: Sat Jan 01, 2011 4:59 pm
 Location: Dhaka
Re: Dhaka Higher Secondary 2010/9
1,2,3,4,5.. Among this numbers , we can form 60 (5P3) three digits numbers... If we put 5 on left, the rest 2 place can be filled by 4P2 > 12 ways and numbers are 512, 513, 514, 521, 523, 524, 531, 532, 534, 541, 542, 543. And among this 12 numbers, value of 543 is the largest.
Having 1 on left, rest 4 digits can form 12 , 3 digits number. Among them 123 is the lowest.
## what do u think? Did i give some logic?
Having 1 on left, rest 4 digits can form 12 , 3 digits number. Among them 123 is the lowest.
## what do u think? Did i give some logic?
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: Dhaka Higher Secondary 2010/9
Still no solution for the first part!
"Inspiration is needed in geometry, just as much as in poetry."  Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: Dhaka Higher Secondary 2010/9
I think I have a solution, the total number of increasing numbers is and decreasing is 20. Here it is:
While we are writing them as decreasing numbers, then either the first one should be 5,4,3... Now there is three cases
First Value five: the second value can be 4,3 or 2 .if the second value is 4 then the third value(3,2,1) can be input in 3P1=3 ways, similarly if the second value is three then the third value can be input in 2P1=2 ways, if the second value is 2 then the third value can be input in only 1 way.. so when the first value is five, the number of decreasing numbers is 3+2+1=6
First value four: By doing the same we can find the number of decreasing numbers is 1+2=3
first value 3: we can do the same and the result is 1..
So the total number of decreasing numbers is 10... do the same for the increasing numbers.. the sum is 20.
While we are writing them as decreasing numbers, then either the first one should be 5,4,3... Now there is three cases
First Value five: the second value can be 4,3 or 2 .if the second value is 4 then the third value(3,2,1) can be input in 3P1=3 ways, similarly if the second value is three then the third value can be input in 2P1=2 ways, if the second value is 2 then the third value can be input in only 1 way.. so when the first value is five, the number of decreasing numbers is 3+2+1=6
First value four: By doing the same we can find the number of decreasing numbers is 1+2=3
first value 3: we can do the same and the result is 1..
So the total number of decreasing numbers is 10... do the same for the increasing numbers.. the sum is 20.

 Posts: 39
 Joined: Sun Mar 30, 2014 10:40 pm
Re: Dhaka Higher Secondary 2010/9
Total no. of numbers formed is 12.
The maximum sum is 888.
The maximum sum is 888.

 Posts: 39
 Joined: Sun Mar 30, 2014 10:40 pm
Re: Dhaka Higher Secondary 2010/9
I made an error. Total no. of numbers formed will be 20, and max. sum is 888.Ragib Farhat Hasan wrote: ↑Sat Jan 16, 2016 8:16 pmTotal no. of numbers formed is 12.
The maximum sum is 888.