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 Post subject: Intersecting CommitteesPosted: Fri, 15 Jun 2012 17:44:51 UTC
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students formed committees. Every of these committees intersect, meaning that they each have at least one member in common (no committees have the same members). Prove that they can form at least one more committee that intersects each of the existing committees.

The difficult part of this problem is utilizing the fact that we have committees each of which intersects.

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 Post subject: Re: Intersecting CommitteesPosted: Fri, 15 Jun 2012 18:09:32 UTC
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rdj5933mile5math64 wrote:
students formed committees. Every of these committees intersect, meaning that they each have at least one member in common (no committees have the same members). Prove that they can form at least one more committee that intersects each of the existing committees.

The difficult part of this problem is utilizing the fact that we have committees each of which intersects.

In fact, every maximal intersecting family on [n] has size 2^{n-1}, so you can form two more committees.

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 Post subject: Re: Intersecting CommitteesPosted: Fri, 15 Jun 2012 19:00:00 UTC
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outermeasure wrote:
In fact, every maximal family on [n] has size 2^{n-1}

How do you prove that?

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 Post subject: Re: Intersecting CommitteesPosted: Sat, 16 Jun 2012 02:49:57 UTC
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rdj5933mile5math64 wrote:
How do you prove that?

Spoiler:
, so every maximal (1-)intersecting family on [n] has size at most .

On the other hand, if is maximal intersecting and has size , then there exists such that . Since is maximal, with (so ) and with , contradicting .

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 Post subject: Re: Intersecting CommitteesPosted: Sat, 16 Jun 2012 02:59:48 UTC
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, rdj5933mile5math64!

I wondered exactly how we form 30 committees with only 6 students.
I began an exhaustive list so I could familiarize myself with on the problem.
And, as a result, I solved the problem by an inelegant method: Brute Force.

Quote:
students formed committees. .Every of these committees intersect,
meaning that they each have at least one member in common.
No two committees have the same members.

Prove that they can form at least one more committee
that intersects each of the existing committees.

Suppose the six students are:

There are five-member committees.

. .

There are four-member committees.

. .

There are three-member committees.
But we can use only the "first ten".

. .

Each of the"second ten" is the complement of a "first ten" set.
. . Hence, they are disjoint.

Take the 6 five-member committees, the 15 four-member committees,
. . and any 9 of the three-member committees.
We will have 30 committees with at least one member in common.

Then the tenth 3-member committee satisfies the problem.

But one which can be used if totally desperate.

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 Post subject: Re: Intersecting CommitteesPosted: Sat, 16 Jun 2012 03:10:10 UTC
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Soroban wrote:
Take the 6 five-member committees, the 15 four-member committees,
. . and any 9 of the three-member committees.
We will have 30 committees with at least one member in common.

Then the tenth 3-member committee satisfies the problem.

But one which can be used if totally desperate.

That is not a solution at all --- you have just listed some example of intersecting sets (incidentally you listed BDE|ACF twice when you want AEF|BCD) but not an exhaustive list. For example, it could be a "dictatorship" where is in every committee, or generally a "junta" (where the membership of in depends only on , some k<n).

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 Post subject: Re: Intersecting CommitteesPosted: Sat, 16 Jun 2012 17:10:01 UTC
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Joined: Wed, 4 Apr 2012 03:51:40 UTC
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Location: Hockeytown aka Detroit
outermeasure wrote:
rdj5933mile5math64 wrote:
How do you prove that?

Spoiler:
On the other hand, if is maximal intersecting and has size , then there exists such that . Since is maximal, with (so ) and with , contradicting .

Very nice! Thanks for the solution!

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