text: Q:
  
  Counting possible configurations: more boxes than elements
  
  How many ways are to place $N/2+1$ 'kids' in $N$ 'chairs'? ($N$ is odd but I have to compute the enter division.) 
  Edit: Maybe it is easier if I define it as: 
  
  I want to place $n$ 'kids' in $2n+1$ 'chairs'. 
  
  A:
  
  Choose seats in $\binom{2n+1}{n}$ ways, and arrange kids in those seats,  
  so $\;n!\binom{2n+1}{n}\;$ arrangements, or written as a permutation, $^{2n+1}P_n$
  
  A:
  
  We have to pick the chairs we use: $\binom{N}{N/2+1}$ ways.
  We have to pick the arrangement of people in the chairs we chose: $(N/2+1)!$ ways.
  Therefore $\binom{N}{N/2+1} \times (N/2+1)! = \frac{N!}{(N/2-1)!}$ ways.
  
  