Taking the biscuit!

Here is a fun solution.

Let's suppose we have $n$ distinct biscuits. How many different ways can we make a sequence of biscuits to eat using any of our $n$ biscuits only once? .

Let $P(n,k)$ denote the number of ways of choosing $k$ biscuits to eat in order. We have a total of $n$ choices for the first biscuit to eat and a total of $n-1$ ways of choosing the second biscuit to eat (sine we ate the first one) and ... $(n - (k-1))$ ways of choosing the $k^{th}$ biscuit. So $P(n,k) = n\times(n-1)\times ...\times (n - (k-1)) = \frac{n!}{(n-k)!}.$

Now since we can choose to eat any $0\le k \le n$ we have to add up all the possible $P(n,k)$ for valid values of $k$ . So let number of ways we can eat $n$ distinct biscuits be denoted $B(n)$ . It follows that $B(n) = \sum_{k = 0}^{k = n} \frac{n!}{(n-k)!}.$

Plugging in $8$ into the formula $B(n)$ gives us $B(8) = 109601$ which is our answer.