Question Given a $t-(v,k,\lambda)$ design $(X,\mathcal{B})$ and a set $U\subset X$ with $|U|=u\leq t$, what is the number of blocks $B\in\mathcal{B}$ such that $B\cap U=\emptyset$?
The answer is: $\lambda {v-u\choose k}/{v-t\choose k-t}$.
How to find this answer?
I know that $b$ the number of blocks in the original design equals: $\lambda {v \choose t}/{k\choose t}$ and that for a $s\leq t$ the design is also a $s-(v,k,\lambda_s)$ design with $\lambda_s=\lambda {v-s\choose t-s}/{k-s\choose t-s}$.
I tried: first thing I tried was calculating $b-\lambda_u$ and rewriting it, but then I realized this gives the number of blocks that do not contain a given $u$-set, which is not the question (in the original question we want to substract from $b$ the number of all the blocks that even contain one point of $U$.
So then I figured I could use the Principle of Inclusion/Exclusion, but it did not work either.
Looking at the expression it seems to me that some sort of Double Counting proof can be applied. ${v-u\choose k}$ counts the number of $k$-sets we can make in the set $X\U$, which has $v-u$ points... But it's not the derived design, because then the blocks would be of size $k-u$.
Any hint/help is highly appreciated.
