$\binom{n}{n+1} = 0$, right? I was looking at the identity $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}, 1 \leq r \leq n$, so in the case $r = n$ we have $\binom{n}{n} = \binom{n-1}{n-1} + \binom{n-1}{n}$ that is $1 = 1 + \binom{n-1}{n}$ thus $\binom{n-1}{n} = 0$, am I right? 

This is asking how many ways you can take $n$ items from $n-1$ items - there are none. So you are correct.
