# A Permutation Quest

• Oct 10th 2012, 04:05 AM
blackjack21
A Permutation Quest
Attachment 25145

The numbers are 1 to 9 (it contain 1 and 9 also) will be placed in to squares as different number each square

the A should be the biggest number in it's current row and column

B and C should be the minimum in only their column

and according this

How many diffrent value the B can take ?

Ps: İf any formula exist there it will be perfect so it can take too much long when you try one by one tactic

Thanks Best Regars....
• Oct 10th 2012, 08:11 AM
johnsomeone
Re: A Permutation Quest
Why not examine all cases where A = 9. That way A is sure to fulfill its maximization requirement of being largest in both its row and column. Then try to think about how big B could be and still be the minimum in its column - but now remembering that the number 9 is already taken.

After you do that, then drop the condition on A and see if you can't get more B solutions. (Hint: I'm pretty sure you won't be able to - every B that works works with A = 9, though that's something you'll have to reason out for yourself.)

I haven't worked it out, but that seems like how it must work. The requirement for C doesn't seem like it would impact the solution at all, because any small value in the top right corner (1st row, 3rd column) that somehow made C a problem could be switched out with larger values in the 2nd column underneath A. In other words, with A = 9 and a choice for column 1, you can always arrange the remaining 5 numbers so that C is a minimum in the 3rd column - just choose C to be the minimum of those 5 remaining numbers.
• Oct 10th 2012, 02:00 PM
awkward
Re: A Permutation Quest
The problem only asks how many values B can take, not how many arrangements are possible in total. So why not concentrate on the possible values of B? There are only 9 possibilities for B, so maybe it's not very hard to work out which ones are or are not possible.

To show that B=1 is a possibility, for example, all you have to do is find one configuration that meets the constraints of the problem and in which B=1.

Some possible values for B are easy to eliminate. For example, can B=9? I don't think so.
• Oct 10th 2012, 02:55 PM
johnsomeone
Re: A Permutation Quest
I didn't intend to suggest examing all permutations, though perhaps my wording of "Why not examine all cases where A = 9" made that unclear. I meant "Why not set A = 9, and then examine the possibilities for B".

To give a summary of my argument:
1) Set A = 9. Then the condition on A is guaranteed satisfied.
2) Every solution with B = x also has a solution with B = x and A = 9, since you could swap 9 and the A value for any solution where A wasn't 9, to get another solution, with the same B value (since easily B isn't 9, so it isn't being swapped), but with A = 9.
3) Consider all ways to make column 1, with numbers {1, 2, ..., 8} available, so that the B condition is satisfied. List the possible B's you get this way. This gives the answer to this problem.
4) You don't need to worry about the conditon on C, because whenever you've populated A=9 and column 1, simply set C = the minimum of the 5 remaining numbers.
• Oct 10th 2012, 04:24 PM
awkward
Re: A Permutation Quest
Sorry, I didn't mean to give the impression that I was criticizing your approach, I merely meant to point out another alternative.
• Oct 11th 2012, 01:23 PM
blackjack21
Re: A Permutation Quest
sorry for late response thanks to both of you for your ideas and guidance
Quote:

Re: A Permutation Quest

I didn't intend to suggest examing all permutations, though perhaps my wording of "Why not examine all cases where A = 9" made that unclear. I meant "Why not set A = 9, and then examine the possibilities for B".

To give a summary of my argument:
1) Set A = 9. Then the condition on A is guaranteed satisfied.
2) Every solution with B = x also has a solution with B = x and A = 9, since you could swap 9 and the A value for any solution where A wasn't 9, to get another solution, with the same B value (since easily B isn't 9, so it isn't being swapped), but with A = 9.
3) Consider all ways to make column 1, with numbers {1, 2, ..., 8} available, so that the B condition is satisfied. List the possible B's you get this way. This gives the answer to this problem.
4) You don't need to worry about the conditon on C, because whenever you've populated A=9 and column 1, simply set C = the minimum of the 5 remaining numbers.

i have solved it with step by step tried B each time with diffrent numbers untill the reach 7 and keep A max C min in their position actually i was ask for a formula or something like that to solve for speed while during exams its take approxmitly 4 min for me i dont know how was it much long for you
in exam you should spend 1 min to this question maximum :D this is madness ? no this iss Turkey
i hate this kind of questions and exam system
but its matter for me you trying to help

Thanks best regards sorry for bad english...