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Dhaka Higher Secondary 2010/9

Posted: Tue Jan 18, 2011 2:06 pm
by BdMO
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?

Re: Dhaka Higher Secondary 2010/9

Posted: Tue Jan 18, 2011 6:29 pm
by leonardo shawon
decreasing order||| 543 > 432 > 321
increasing order || 123 < 234 < 345

the sum is 888

Re: Dhaka Higher Secondary 2010/9

Posted: Tue Jan 18, 2011 7:22 pm
by Moon
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:
what if we forget that the digits can be increasing or decreasing?

Re: Dhaka Higher Secondary 2010/9

Posted: Wed Jan 19, 2011 12:06 am
by leonardo shawon
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?

Re: Dhaka Higher Secondary 2010/9

Posted: Tue Feb 01, 2011 11:23 pm
by Moon
Still no solution for the first part!

Re: Dhaka Higher Secondary 2010/9

Posted: Sat Aug 20, 2011 6:20 pm
by Shifat
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. :?:

Re: Dhaka Higher Secondary 2010/9

Posted: Sat Jan 16, 2016 8:16 pm
by Ragib Farhat Hasan
Total no. of numbers formed is 12.
The maximum sum is 888.

Re: Dhaka Higher Secondary 2010/9

Posted: Wed Aug 29, 2018 7:55 pm
by Ragib Farhat Hasan
Ragib Farhat Hasan wrote:
Sat Jan 16, 2016 8:16 pm
Total no. of numbers formed is 12.
The maximum sum is 888.
I made an error. Total no. of numbers formed will be 20, and max. sum is 888.