Find the number of ways in which $n$ distinct balls can be distributed between $n$ children such that exactly one child doesn't get a ball.
This can be solved easily by looking at it as division and distribution.
Here, $n-2$ children get exactly $1$ ball, one child gets $2$ balls and one child gets no ball. The number of possible divisions is just
$$\frac{n!}{0!2!(1!)^{n-2}(n-2)!}=\frac{n!}{(n-2)!2!}$$
Hence the number of possible distributions is
$$\frac{n!}{(n-2)!2!}\times n!=^nC_2\times n!$$
This answer looks suspiciously close to the answers we would get by using the normal selection and permutation method. However I couldn't figure out any "normal" way. Can you help me out?
Make $n$ piles with $0,2, 1,1,1...$ balls with the pile of $2$ chosen in $\binom n 2$ ways, and the piles permuted in $n!$ ways giving $\binom n 2 n!$ arrangements
Line up the kids in (say) alphabetical order and distribute.