ISI 2004 MIII

Question

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

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

Then $x$ has 

• A. exactly two values both between $4$ and $8$.
• B. exactly one value and this lies between $4$ and $8$.
• C. exactly two values both between $8$ and $16$.
• D. exactly one value and this lies between $8$ and $16$.

Comments

yes i also  getting 6

we can just realize it as the number of edges in complete graph of order n=n(n-2)/2.

or in other way we can say that if there are n stations that are mutually connected with each other then there will be n(n-1)/2 total keys in channel(mesh topology)

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Explain how and in detail

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it is no of edge in complete graph

Answer 1

(B) is ans.

Comments

I am getting A is answer

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wt is the another value u r getting behalf of 6?

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4 and 6

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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.

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Sorry i made an mistake you are right

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values are 4,5 and 6 all are satisfying the condition.Ans should be A.

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Prove it

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

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good explanation @2018 loyal

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By going through the options. Yes B is the answer but I think 7, 8 will never be possible.

Am i right??

Can someone clarify this please?

Answer 2

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.

Comments

4C2 ??? Choosing 2 groups out of 4 groups ... how does it ensure that 2 groups has 1 member in common ...

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@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.

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Yes ...exactly

Answer 3

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

Comments

explanation??