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A club with $x$ members is organized into four committees such that

(a) Each member is in exactly two committees

(b) Any two committees have exactly one person in common.

Then $x$ has

(1) exactly two values both between $4$ and $8$

(2) exactly one value that lies between $4$ and $8$

(3) exactly two values both between $8$ and $16$

(4) exactly one value that lies between $8$ and $16$ .

Which one/one of this options are correct ?

I ran out of all ideas . Help needed . Complete solution will be highly appreciated .

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Conditions a) and b) mean that there is a bijection from people to (unordered) pairs of committees. So how many pairs of committees are there?

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  • $\begingroup$ There are $\frac {4 \times 3}{2}=6 $ , right ? $\endgroup$ – Suman Kundu May 8 '17 at 15:03
  • $\begingroup$ Yes, that's right. $\endgroup$ – Especially Lime May 8 '17 at 15:15
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Hint:

consider a complete graph on four points $K_4$ and what the edges and vertices represent here.

enter image description here

Then consider how to add more members satisfying (a) and (b)

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