One method for threshold secret sharing involves splitting a secret into multiple parts and distributing it amongst $n$ people such that any subset of threshold $t$ members can reconstruct the secret, but no combination of $(t - 1)$ or fewer can do so.
In order to ensure that any combination of $t$ members is guaranteed to have all parts of the secret, each part must be distributed to at least $(n - t + 1)$ people. It is inefficient to distribute 2 different parts to the same subset of people. There are $n\choose{n - t + 1}$ ways of distributing a part to this number of people and I have noticed that, for any $n$ and $t$, this is the same as the number of parts required to ensure no subset with $(t-1)$ can reconstruct the secret. In other other words there is exactly one way of doing this efficiently.
I feel it should be simple to show why this is the case, but I'm very rusty and I can't see it. Is there a simple proof?