There is not a $2-(21,6,1)$ design!

Question

I've been asked to prove There is not a $2-(21,6,1)$ design.

for a given $2-(v,k,\lambda )$ design assume that $r$ is the number of blocks containing a given point and $b$ is the number of blocks, we know that $r(k-1)=\lambda (v-1)$ and $bk=rv$ but these didn't lead to contradiction because $r$ and $b$ both seemed to be integers...

on another hand I know that there exists a $2-(21,5,1)$ design I thought maybe we can say if there exists a $2-(v,k,\lambda )$ design then there will be no $2-(v,k+1,\lambda )$ design.

Answer 1

each block has $\binom{6}{2}=15$ pairs, the total number of pairs is $\binom{21}{2}=210$, so there must be $14$ sets.

We conclude each element belongs to $\frac{6\times 14}{21}=4$ subsets.

It follows that the elements that contain point $x$ only intersect pairwise at $x$. ( since the $4$ subsets that contain $x$ must cover all other elements).

Therefore any two subsets intersect at at most one point.

We can now obtain a contradiction. Let $A_1,A_2,A_3,A_4$ be the four sets that contain $x$. Notice that any other subset must intersect one of these four sets in more than one point. A contradiction.