Solve 17 Difficult Math Problems: Proving At Least 2 Committees are Identical

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Discussion Overview

The discussion revolves around a combinatorial problem involving four individuals forming 17 committees to tackle 17 math problems. Participants are tasked with proving that at least two of these committees must consist of the same members, exploring various counting methods and reasoning to arrive at this conclusion.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant suggests using the rule of product to count the number of ways to form groups from the four individuals, implying that each person can either belong to or not belong to a committee.
  • Another participant reiterates the problem and presents a table to illustrate the 15 possible unique committees that can be formed from the four individuals, concluding that at least two committees must be identical given the formation of 17 committees.
  • A different approach is proposed where the participant counts the number of committees based on the number of members (1, 2, 3, or 4), calculating the total to be 15 unique committees.
  • Another participant confirms the total number of possible committees as 16 (including the empty committee) and notes that excluding the empty committee leaves 15 valid committees.

Areas of Agreement / Disagreement

Participants generally agree that there are 15 possible unique committees that can be formed from the four individuals. However, the discussion includes various methods of counting and reasoning, indicating that while the conclusion about the necessity of identical committees is accepted, the approaches to reach that conclusion vary.

Contextual Notes

Participants rely on combinatorial reasoning and the counting of subsets, with some methods involving direct enumeration and others using combinatorial formulas. The discussion does not resolve the nuances of each counting method or the implications of the results.

yakin
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There are 4 people in a class. Among them, they are to solve 17 difficult math problems.

They form 17 committees – one committee to deal with each of the 17 problems.

Prove that at least 2 of these committees contain exactly the same people.

PLEASE EXPLAIN HOW EACH STEP TO REACH TO THE CONCLUSION.
THANKS!
 
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How many groups can be formed from these four people? Hint: Obviously, each person can belong or not belong to a group. Then use the rule of product to count the number of ways to include/exclude all four people.
 
Hello, yakin!

There are 4 people in a class. Among them,
they are to solve 17 difficult math problems.

They form 17 committees – one committee
to deal with each of the 17 problems.

Prove that at least 2 of these committees
contain exactly the same people.
With 4 people, there are only 15 possible committees.

. . \begin{array}{ccccc} 1&A&B&C&D \\ 2&A&B&C&- \\ 3&A&B&-&D \\ 4&A&B&-&- \\ 5&A&-&C&D \\ 6&A&-&C&- \\ 7&A&-&-&D \\ 8&A&-&-&- \\ 9&-&B&C&D \\ 10&-&B&C&- \\ 11&-&B&-&D \\ 12&-&B&-&- \\ 13&-&-&C&D \\ 14&-&-&C&- \\ 15&-&-&-&D \\ \cancel{16}&-&-&-&- \end{array}

Therefore, at least two of the 17 committees
. . must contain exactly the same people.
 
This is how I look at it: suppose we want to make as many different committees as possible.

We have 4 possibilities: a committee may contain 1,2,3, or 4 people.

There is only ONE way to make a committee of 4: all the students belong to it.

There are 4 ways to make a committee of one, one way for each student.

There are 4 ways to make a committee of 3 (since that amounts to choosing "who to leave out").

So we are left with the final possibility of how many ways we can choose a committee of 2 students. This is given by "4 choose 2" or:

$\dfrac{4!}{2!(4-2)!} = \dfrac{24}{2\ast2} = \dfrac{24}{4} = 6$.

Adding up these possibilities we get:

1+4+4+6 = 15 possible DIFFERENT committees.

Since we must choose 17 committees, 2 of these must match (individually) one of the 15 possible (although more matches might actually occur).
 

Another way to count the possible cases . . .

There are four people available.

For each person, we have two choices:
. . (1) Yes (on the committee),
. . (2) No (not on the committee).

Hence, there are: 2^4 \,=\,16 possible choices.

But this list includes No-No-No-No
. . (no one is on the committee).

Therefore, there are 15 possible committees.
 

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