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A club with $n$ members is organized into four committees so that each member belongs to exactly two committees and each pair of committees has exactly one member in common. Then

  1. $n = 4$
  2. $n = 6$
  3. $n = 8$
  4. $n$ cannot be determined from the given information

4 Answers

Best answer
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Let the first committee $=  \{1,2,3\}$

Second $=\{3,4,5\}$

Third $= \{5,6,1\}$

Fourth $=\{2,4,6\}$

$n=6$

Now, if we take intersection of any two committees we will get only one member in common, and also each member is part of exactly two groups.

We can also imagine it as an graph theory problem. $4$ committees $= 4$ nodes and solution is to draw edges such that each edge is part of exactly two nodes and answer we will get is complete graph with $4$ nodes (having total six edges.)

Hence, answer B

edited by
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$\text{There are 4 committees, assuming}$ $\color{blue}{a,b,c,d}$.

$\text{Now, each member is in}$ $\color{red}{\text{exactly 2 committees}}$.

$\text{possible combination of 2 committees out of 4 committees will be}$ $\color{violet}{^4C_2 = 6}$ i.e. $\color{maroon}{\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\},\{c,d\}}$

$\text{And this was also the criteria that}$ $\color{red}{\text{any two committees have exactly one member in common}}$.

$\text{And as we've already seen that the}$ $\color{green}{\text{possible combination of any two committees is 6, then the number of members also be}}$ $\color{orange}{6}$.
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(B) n = 6

Because, if we make 4 divisions (c1,c2,c3,c4), we can only take 3 members into one committee if there is only overlapping of one person.  

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