# Ways to award first and second place to two persons out of nine

Question: In how many ways can the first and second place be awarded to two persons from among 9 people.

my answer is = 9!/(9-2)!2!
= 9!/7!2!
= 36


however the suggested answer I found on the book is

  9!/7! which is 72


First choose a person to get the first place. There are $9$ candidates. Then choose one to get the second place. There are $8$ candidates left. This leads to $9\times 8=72$ possibilities. Things would have been different if there had been no distinction among the two persons picked out. Naming them $A$ and $B$ the possibilities $(A,B)$ and $(B,A)$ would stand for the same selection. The double counting must be repaired then by dividing by $2$ giving $36$ as outcome.
In general if $k$ persons are picked out and each of them gets a special position then there are $$n\times\left(n-1\right)\times\cdots\times\left(n-k+1\right)=\frac{n!}{\left(n-k\right)!}$$ possibilities.
If $k$ persons are picked out that do not get a special position then there are $$\binom{n}{k}=\frac{n!}{k!\left(n-k\right)!}$$ possibilities. Here factor $k!$ somehow repairs the 'more than once' counting.