I'd like to help with the following problem:
$$ \binom{x}{r-1} + \binom{x}{r} = \binom{x+1}{r} \tag{8.6}\label{8.6} $$ 7. Let $A(r, n)$ be the number of distributions leaving none of the n cells empty. Show by a combinatorial argument that $$ A(r, n+1) = \sum_{k=1}^r \binom{r}{k} A(r-k, n) \tag{11.5}\label{11.5} $$ Conclude that $$ A(r, n) = \sum_{v=0}^n (-1)^v \binom{n}{v} (n-v)^r \tag{11.6}\label{11.6} $$ Hint: Use induction; assume $\ref{11.6}$ to hold and express $A(r-k,n)$ in $\ref{11.5}$ accordingly. Change the order of summation and use the binomial formula to express $A(r, n+1)$ as the difference of two simple sums. Replace in the second sum $v+1$ by a new index of summation and use $\ref{8.6}$.
According to the Hint:
$$ \begin{eqnarray} A(r, n+1) &=& \sum_{k=1}^r \binom{r}{k} \sum_{v=0}^n (-1)^v \binom{n}{v} (n-v)^{r-k} \\ &=& \sum_{v=0}^n (-1)^v \binom{n}{v} \sum_{k=1}^r \binom{r}{k} (n-v)^{r-k} \\ &=& \sum_{v=0}^n (-1)^v \binom{n}{v} (n-v+1)^r \tag{*}\label{*} \\ \end{eqnarray} $$
I'm now stuck on this.
I've tried to evaluate the $A(r,n)$ for $n=4$ and $n=5$, and noticed that the expression could perhaps be split by the value of $(-1)^v$:
$$ A(r, n) = \sum_{v=0}^{\lfloor n/2 \rfloor} \binom{n}{2v} (n+1-2v)^r - \sum_{v=0}^{\lfloor n/2 \rfloor} \binom{n}{2v+1} (n-2v)^r $$
But this seems to be wrong because of the floor function and I don't know where to go from there either. Into which simple sums can I split $\ref{*}$ please?
There are other question about this number but I'm interested only in the bold part in the quote.
Here's your mistake: $$ \sum_{v=0}^n (-1)^v \binom{n}{v} \sum_{k=1}^r \binom{r}{k} (n-v)^{r-k} = \sum_{v=0}^n (-1)^v \binom{n}{v} [(n-v+1)^r-(n-v)^r], $$ because in the binomial formula the index runs from zero to $r$ while here it starts at $k=1$, so you have to subtract the $k=0$ term.