I have a collection of flavors being sampled in batches of $3$. I know that each pair of flavors occur together in exactly one batch. Also each flavor appears in the same number of batches. How can i prove that there are a total of either $3$, $6n+1$ or $6n+3$ flavors?
So i know that this can be solved by using combinatorial design theory. So i have knowledge about thsi kind of desings:
$(v,k,\lambda)$ and t-designs: $(v,k,\lambda_{t})$
and basically i have normally came across problems with 3 given data, and here i only have two: $v=?$ $k=3$ $\lambda_{2}=1$ I am not completely sure, if it's a 2-design or a design. I think that probably i should use some of the basic formula and also my knowledge of these subject is not really advanced, so i have only the recursion for $\lambda$ and two or three other formulas
Any help would be appreciated. Thank you in advance
You know that there are ${v \choose 2} = v(v-1)/2 = \frac{1}{2}(v^{2}-v)$ total pairs of flavors, and each must show up exactly once in a batch. Since each batch has size $3$, each of the $b$ batches has ${3 \choose 2} = 3$ pairs showing up. So $3b = \frac{1}{2}(v^{2}-v)$, or $6b = v^{2}-v$.
Now you have a quadratic equation, $v^{2}-v - 6b = 0$. Find the solutions in terms of $b$ using the quadratic formula. Keep in mind they have to be positive integers (since they are the number of flavors). Better yet, notice that this equation implies that $v^{2}-v \equiv 0 \bmod{6}$, and use this to show that $v \equiv 1$ or $3 \bmod{6}$. In this case you will need an extra argument to show that $v \not\equiv 0 \bmod{6}$.