How many solutions are there to $x_1 + x_2 + x_3 + x_4 = 30$ s.t. $x_1 + x_2 le 20$ and $x_3 ge 7$?

Here is the original question from the book:

How many ways are there to distribute $30$ green balls to $4$ persons if Alice and Eve together get no more than $20$ and Lucky gets at least $7$?

I rewrote the problem as the title of this post, i.e. how many solutions are there to $x_1 + x_2 + x_3 + x_4 = 30$ such that $x_1 + x_2 le 20$ (Alice and Eve together get no more than $20$) and $x_3 ge 7$ (Lucky gets at least $7$).

First, I "eliminated" the second restriction. Give Lucky $7$ balls and distribute the remaining $23$. Thus, we have

$$
x_1 + x_2 + (x_3 + 7) + x_4 = 30 \
x_1 + x_2 + x_3 + x_4 = 23
$$

Then, since the other restriction left me with doubts, I counted the number of solutions to the equation above, which is $23+3 choose 3$, and tried counting the solutions that violate $x_1 + x_2 le 20$ to subtract from $23+3 choose 3$.

Let $k = x_1 + x_2 le 20$. The solutions that violate this are those where $k ge 21$. Hence

$$
(k + 21) + x_3 + x_4 = 23 \
k + x_3 + x_4 = 2
$$

which has $2+2 choose 2$ solutions. Therefore, my solutions was ${23+3 choose 3}-{2+2 choose 2} = 2600 – 6 = 2594$.

However, the book provides the following solution:

Suppose Alice and Eve together get $k$ balls. This can be done in $k + 1$ ways. That leaves $30 – k$ balls for the other $2$ persons, but Lucky must get at least $7$ of these. So, there are $30 – k – 7$ additional balls to distribute to Lucky and the fourth person. This can be done in $(30 – k – 7) + 1$ ways. Hence, our answer is $sum_{k=0}^{20}(k+1)(24-k)$.

That gives $2464$. Forgive me if I didn’t get some "obvious" detail, but what the heck does that summation mean? I was getting everything until it appeared, although I do see what the summands are. Could you point out where is the mistake in my solution? If you think the answer provided is simpler, please do explain.

Thank you very much for any clarifications!