I am having trouble with the wording of this question.
$Tn$ is the number of derangements of a set of size $n$.
$n$ treeloppers are cutting down $n+1$ trees.
They go and have a lunch break in the lodge and return to start chopping the $n+1$ trees again.
How can I show that there are $Tn$ + $T(n+1)$ ways that they can chop the trees so that no one is chopping the same tree as they were before lunch?
I am approaching this problem by trying to let the equation specifying the number of permutations ${1,2,...n}$ with exactly $k$ fixed points equal to the addition of the derangement formula for $n$ and $n+1$. I am fairly confident that this is not the correct approach as, the question seems too simple for this method.
a)$$n!\sum_{r=0}^{n}(-1)^r/r! = T_n$$
b)$$n+1!\sum_{r=0}^{n+1}(-1)^r/r! = T_{n+1}$$
c)$$\frac{n!}{k!} \sum_{r=0}^{n-k}(-1)^r/r!$$
And trying to get $a+b$ to equal $c$.
How should I be attempting this problem?
Hint:
After lunch there are two cases: (1) The same tree that wasn't being chopped before lunch is still not being chopped; and (2) A different tree after lunch isn't being chopped.