Filling at least $1$ box with a white ball from $1$ to $n$ numbered boxes 
Given $n$ boxes numbered $1$ to $n$, each box is to be filled with either a white ball or a blue ball
such that at least one box contains a white ball and boxes containing white balls are consecutively
numbered.  What is the total number of ways this can be done?


Solution provided:
The task is similar to choosing two out of the $n + 1$ crosses to mark the start and
end of the consecutively numbered boxes that contain white balls.
This is ${n+1 \choose 2}$, which is equal to $\frac{n(n + 1)}{2}$.
My question:
I get that the picture shows $n$ boxes because we can have $1$ to $n$ white balls.
I have a feeling they are using the $r$-combination formula here: ${r + n -1 \choose r}$.
That's all about I get. I have no clue why they have $n+1$ crosses or why they are choosing only $2$ from those crosses or what those crosses represent even?

PS: I know there is sum of sequence approach to this but I'm required to provide a combinatorial working.
 A: To make a $7$-lettered string $$BBBBWWW$$ given the $7$ slots
$$ \square \square \square \square \square \square \square $$
first separate the squares where consecutive $W$'s are to be placed like this
$$ \square \square \square \square \times \square \square \square \times $$
and then place the W's into squares between crosses (and B's in rest).
The two crosses, which mark the beginning and end of substring of $W$'s, can be placed in any $n+1$ places around $n$ slots, hence $$ \binom{n+1}{2}$$
A: The question requires a single string of consecutive boxes to contain white balls, while all the remaining boxes contain blue balls. As an example, if there are ten boxes, a possible situation satisfying the conditions would be BBWWWWWBBB.
The provided solution is saying that we can count this by just choosing the first and last box(es) that contain the white balls. In the above example, we would choose boxes 3 and 7. Thus the number of ways of filling the boxes would be equivalent to the number of ways of choosing the first and last box.
Normally, that would just be $\binom{n}{2}$. However in this case, the conditions specifically allow for just a single box to be white, and this would not be accounted for by $\binom{n}{2}$ as the first and last boxes would be the same box. That is why they threw in those little x's.
Now for the example I gave above, the configuration would be xBxBxWxWxWxWxWxBxBxBx; instead of choosing the first and last boxes, I choose the x before the first box with a white ball and the x after the last box with a white ball, i.e. the third and eighth x. By doing it this way, we can choose a single box to contain a white ball, by just choosing the x's right before and right after that single box. But since there will always be one more x than there are boxes, the number of ways will now be $\binom{n+1}{2}$.
Note that another simple way to get around the problem of single boxes would just be to add the number of ways of choosing a single box ($n$) to the earlier $\binom{n}{2}$ and get
$$ \binom{n}{2} + n = \frac{n^2-n}{2} + n = \frac{n^2+n}{2} = \binom{n+1}{2} $$
