32. $[\mathbf{1 7}]$ How many 48-tuples of positive integers $\left(a_{1}, a_{2}, \ldots, a_{48}\right)$ between 0 and 100 inclusive have the property that for all $1 \leq i<j \leq 48, a_{i} \notin\left\{a_{j}, a_{j}+1\right\}$ ?
Answer: $54^{48}$
(With Ashwin Sah) The key idea is write the elements of the sequence in increasing order. These sets are in bijection with solutions to $d_{1}+\ldots+d_{k}=48$ and $a_{1}+\ldots+a_{k+1}=53$ with $d_{i} \geq 1, a_{i} \geq 1$ for $2 \leq I \leq k$, and $a_{1}, a_{k+1} \geq 0$. Notice that there are $\binom{54}{k}$ solutions to the second equation and then there are $\frac{48!}{d_{1}!\cdots d_{k}!}$ solutions for each $\left\{d_{i}\right\}$ set. Then this gives that the answer is
$$
\begin{array}{c}
\sum_{1 \leq k \leq 48}\binom{54}{k} \sum_{d_{1}+\ldots+d_{k}=48} \frac{48!}{\prod_{i=1}^{k} d_{i}!} \\
=48!\left[x^{48}\right] \sum_{1 \leq k \leq 48}\left(e^{x}-1\right)^{k}\binom{54}{k} \\
=48!\left[x^{48}\right] \sum_{0 \leq k \leq 54}\left(e^{x}-1\right)^{k}\binom{54}{k} \\
=48!\left[x^{48}\right]\left(e^{x}\right)^{54} \\
=54^{48} .
\end{array}
$$