The way I like to is to procede the opposite.Originally Posted by Jameson
Meaning do the problem:
Then from the full total subtract this number.
I do not understand why you chose to place this Algebra section.
I am studying for the Putnam competition in December and I want just a hint at how to do this problem, not a full solution please. I'm trying to develop a good methodology to approaching proofs.
A1. Determine, with proof, the number of ordered triples of sets which have the property that
(i) , and
(ii) . Express the answer in the form of , where a,b,c, and d are non-negative integers.
What's the best way to approach this?
Ok. I'll try to start this way. My problem is that I have trouble going from specific cases and setting up a pattern/generalizing behavior that's needed for a proof.Originally Posted by ThePerfectHacker
So if , then obviously the sets can not have any items in common. So I need to find the number of ways to split up the numbers 1-10 into 3 groups.
I see this as having , , and , where i+j+k=10. On the right track or should I stop and consider another method? If I'm going in the right direction, any suggestions after the step I gave?
Thanks guys.
PH - Are you going to take the Putnam exam?
Sorry, I meant to write.
(Still number theory is not appropriate. I would place it in the High School Probabily question. Yes I know this is very complicated but as far as level is concerned it deserves to be there. I guess I am the only one of the forum who is careful where everything goes).
I solved this problem and got the answer which does not have the required form. I found the reason for this is that the problem is not stated correctly: there is an equal sign in the second condition.Originally Posted by Jameson
Your methodology so far is off track. Look carefully at what conditions (i) and the correct (ii) mean. This is not a problem of how to place 10 objects in 3 containers. Up to 30 objects could be used.
Ok, I finally got it.
Drawing a Venn Diagram with three circles there are seven distinct regions. From condition 1 all of these regions may be used, but because of condition 2 the center region is excluded.
So there are 6 regions which we can place 10 elements in. Thus the answer is ways, but in the form they requested