Assume that $2\leq k <v$ and $\lambda>0$ and let $S=\{1,2,...,v\}$. We know that $A_1,A_2,...,A_b$ are subsets of $S$ for which $|A_i|=k,\; \forall i$. Prove that if any two of the three conditions hold then the third one also holds and the sets $A_1,A_2, . . . , A_b$ form a $(v, k, λ)$-design:
(i) $\displaystyle b=\frac{ λv(v − 1)}{k(k − 1)}$
(ii) Every subset of size $2$ of the set $S$ is included in the set $B_i$ for at least $λ$ indices $i$.
(iii) Every subset of size $2$ of the set $S$ is included in the set $B_i$ for at most $λ$ indices $i$.
My attempts: If (ii) and (iii) hold, the sets by definition of a 2-design form a $(v,k,\lambda)$-design. Then we know that the basic equations $r(k-1)=\lambda(v-1)$ and $bk=vr$ hold so it is easy to solve for $b=\frac{ λv(v − 1)}{k(k − 1)}$.
Then to prove the other 2 cases, I defined a matrix $\boldsymbol{M}_{\binom{v}{2}\times b}$ such that we name the rows as all the subsets (of size 2) of the set $S$ and that the place $t$ on the row corresponding to the subset $\{i,j\}$ is $1$, if ${i, j} \subseteq B_t$ and zero otherwise. I want to first show that the number of $1$'s in the matrix is $b\binom{k}{2}$. I don't know how to finish the arguments so can someone do that or prove these some other way?