How many ways can I sort 50 distinct items into 2 lists with no repetition, order matters?

Question

I must use all $50$ items, but either list can be empty.

I know that the default answer is $2^k$ for $k$ elements when order does not matter. However, I am not sure how to arrive at the answer when order does matter. (As an aside: if there were three lists, would it be $3^k$ if order did not matter?)

I have attempted to break down the problem in the title as $3$ items into $2$ lists. The total number of combinations is $24$, but I am still unsure of how to extrapolate that number into a more general form. I have looked into a "stars and bars" solution but that doesn't give me $24$ either.

The answer appears to be $4!$ for $3$ items, but I have only vague ideas as to why. Appealing to the multiplication principle with repetitions forbidden, I proposed that we have $4$ boxes, where the first box can be either the first, second, third, or none of the items. Is this the correct way to think about it? If so, is $51!$ the answer when there are $50$ items?

Answer 1

How many ways can you arrange 51 objects? The 50 items, plus a divider that splits the items into two lists?

Answer 2

Let's say you have $0 \leq n \leq 50$ items on the first list.

First, choose which $n$ items you want on the list ($_{50}C_n$ possibilities). Then choose what order you want them in ($n!$ possibilities).

You've already chosen what goes on the other list (both how many, and which ones). So, choose the order of those items on the other list ($(50-n)!$ possibilities).

Then, the magic happens when we sum over $n$:

$$N = \sum_{n=0}^{50}\frac{50!}{n!(50-n)!}n!(50-n)! = \sum_{n=0}^{50}50! = 51 \cdot 50! = 51!.$$