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+8 votes

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

  1.  each member is in exactly two comittees,
  2.  any two committees have exactly one member in common .

Then $x$ has 

  1. exactly two values both between 4 and 8.
  2. exactly one value and this lies between 4 and 8.
  3. exactly two values both between 8 and 16.
  4. exactly one value and this lies between 8 and 16.


asked in Combinatory by Veteran (15k points)
edited by | 222 views
I am getting A as answer can some one verify.
i am getting 6 as answer.what other value are u getting except 6??

3 Answers

+10 votes
Best answer

B is ans.

answered by Boss (7.2k points)
selected by
I am getting A is answer
wt is the another value u r getting behalf of 6?
4 and 6
can u show your relation bcz given 2 points a and b are the condition of complete graph.
where each edge can share exactly 2 committees mean any of the two committees has exactly one edge in comman.
Sorry i made an mistake you are right
values are 4,5 and 6 all are satisfying the condition.Ans should be A.
Prove it

Sorry it's my wrong interpretation , i understood exactly as at least

0 votes
every two group has exactly 1 member common so no. of common members between each two groups among the  four groups = 4c2 = 6    

these six members are distinct from each other because it is given that each member is in exactly two committees .      so ans is 6.
answered by Junior (717 points)
4C2 ??? Choosing 2 groups out of 4 groups ... how does it ensure that 2 groups has 1 member in common ...

@Puja  Mishra. Suppose 4 committees are A,B,C and D .
Now suppose there is a member of the club, say Y. Now he has to choose 2 committees out of 4. 
In how many ways he can choose? 4C2 = 6 ways.
1)AB  2)AC 3)AD 4)BC 5)BC 6)CD
Y can choose any 1 of these combinations. Suppose he chooses AB. So AB has a member in common- Y. Both the conditions are satisfied. 
Now suppose there is another member of the club, say Z. Z can't choose combination AB.Choosing AB will violate condition (ii). In this way, Each [possible combinations of the committee has a different member in common.

–1 vote

The correct answer is 6. Hence, option (B).

answered by Veteran (16.5k points)

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