Let $v\in \mathbb{N}$ and $v\geq3$ and $b\leq v$
I want to prove that there exist a $2-design$ with parameters $(v,v-1,v-2)$ Also i need to evaluate if there exist a $2-design$ with parameters $(v,v-1,\lambda_{2})$ where $\lambda_{2} \neq v-2$
So for the prove i need to check those different divisors(for s: $0\leq s \leq t-1$, if it's any t-design), but since it's a $2-design$ it's just two options, $s$ is either $0$ or $1$. Or am i wrong, but i am confused how to check it. There aren't many formulas there.
But for the second part, should i just check for that example, or not ?
Any help would be appreciated.
The $(v, v-1, v-2)$ design on the points $X=\{1,2,\dots,v\}$ is given by $\{X\backslash\{1\}, X\backslash\{2\}, \dots, X\backslash\{v\}\}$.
If the design's $\lambda$ is allowed to vary, while keeping $v$ and $k=v-1$ constant, we note that in the fundamental relation of block designs $$\lambda(v-1)=r(k-1)=r(v-2)$$ $v-1$ and $v-2$ are coprime, implying that $\lambda=x(v-2)$ and $r=x(v-1)$ for some $x\in\mathbb N$. A $(v, v-1, x(v-2))$ design can only be the $(v, v-1, v-2)$ design repeated $x$ times, but such a design is not simple for $x>1$.
To summarise, if repeated blocks are allowed, $\lambda_2=x(v-2)$ for $x\in\mathbb N$. If the design is restricted to be simple, there are no solutions beyond $\lambda_2=v-2$.