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How many different (7,7,3,3,1) BIBD are there on vertex set[7]? Note that two such designs are different if their sets of blocks are different. That is, we do not require that the BIBDs be non-isomorphic; we simply require that they be non-identical.

I have found that we use construct the fano plane shape to illustrate this problem but counting part is difficult. I found the two ways to construct fano plane but any suggestion on how to count them up?

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I don't understand this question. You have a unique such object upto isomorphism---Fano plane. Are you looking for the number of ways to renumber the 7 points on the Fano Plane? –  user21436 Apr 19 '12 at 6:19
    
@Kannappan, I understand the problem to be, given a 7-set, how many different Fano planes are there, if Fano planes are different when they don't have the same 7 3-sets. See my answer. –  Gerry Myerson Apr 19 '12 at 6:33
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1 Answer

Let the vertices be 1, 2, 3, 4, 5, 6, 7. The blocks containing 1 must be 1ab, 1cd, and 1ef, where $\{{\,a,b,c,d,e,f\,\}}=\{{\,2,3,4,5,6,7\,\}}$. So the first question is, how many ways can you partition a 6-set into 3 2-sets?

Now, the block containing a and c must contain either e or f. Once you have decided which, the rest is forced, e.g., if ace is a block, the other blocks must be adf, bcf, and bde.

That should give you all you need.

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Oh, I understand this now. Thank you. +1. So, OP considers two designs different if they have different block sets, then? –  user21436 Apr 19 '12 at 6:35
    
Ah, yes. Right. D'oh! –  user21436 Apr 19 '12 at 6:36
    
the number of ways to partition a 6-set into 3 2-sets are 6C2*3!, and guess we need to subtract the overlaps, but not sure the line you mentioned about the the rest is forced. –  user28699 Apr 19 '12 at 8:38
    
Partitioning the 6-set; $a$ must go with something, and there are 5 options for that something. Suppose $a$ goes with $b$; then $c$ must go with something, and there are 3 options for that something. Once you decide what $c$ goes with, there is only one option for what to do with the remaining two things - they must go with each other. As for the forced part, if you already have blocks 1ab, 1cd, 1ef, and ace, then a has appeared with 1, b, c, and e, so there must be a block adf; c has appeared with 1, d, a, and e, so there must be a block bcf; etc. –  Gerry Myerson Apr 19 '12 at 13:04
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