Solution:

Let the numbers of marbles that Alex, Billy, and Charles get be $a$, $b$, and $c$, respectively, with $a$, $b$, $c \geq 0$ and $a + b + c = 7$. We require that $a$, $b$, and $c$ are all distinct (no two are equal).

Since the marbles are identical and the people are distinguishable, we are counting ordered triples $(a, b, c)$ of non-negative integers, all distinct, with sum $7$.

First, list all possible unordered triples of distinct non-negative integers that sum to $7$.

Let $a < b < c$ (since the variables are symmetric, we can count the number of such ordered triples for each unordered triple by multiplying by $6$ for all permutations).

Let us find all sets $\{a, b, c\}$ with $a < b < c$, $a, b, c \geq 0$, $a + b + c = 7$.

Try $a = 0$:

Then $b < c$, $b \geq 1$, $c \geq b + 1$.

$0 + b + c = 7 \implies b + c = 7$

So $b$ can be $1, 2, 3$:

- $b = 1$, $c = 6$
- $b = 2$, $c = 5$
- $b = 3$, $c = 4$

So possible triples: $(0, 1, 6)$, $(0, 2, 5)$, $(0, 3, 4)$

Try $a = 1$:

Then $b < c$, $b \geq 2$, $c \geq b + 1$.

$1 + b + c = 7 \implies b + c = 6$

So $b = 2, 3$:

- $b = 2$, $c = 4$
- $b = 3$, $c = 3$ (but $b = c$ not allowed)

So only $(1, 2, 4)$

Try $a = 2$:

Then $b + c = 5$, $b \geq 3$, $c \geq b + 1$

- $b = 3$, $c = 2$ (not possible, $c > b$)
- $b = 4$, $c = 1$ (not possible, $c > b$)

So no solutions.

Try $a = 3$:

Then $b + c = 4$, $b \geq 4$, $c \geq b + 1$

But $b = 4$, $c = 0$ (not possible, $c > b$)

So no solutions.

Thus, the possible unordered triples are:

$(0, 1, 6)$
$(0, 2, 5)$
$(0, 3, 4)$
$(1, 2, 4)$

Each unordered triple corresponds to $3! = 6$ ordered triples (since all numbers are distinct).

So total number of ways is $4 \times 6 = 24$.

Answer: $\boxed{24}$