How many ways can 25 red balls be put into 3 distinguishble boxes if no box is to contain more than 15 balls? I'm reading Martin's: Counting: The Art of Enumerative Combinatorics.

How many ways can 25 red balls be put into 3 distinguishble boxes if no box is to contain more than 15 balls?

I understand that it's the solutions to the equation $x+y+z=2$5 minus the solutions to the equation $x+y+z=25-16$. The answer in the book is:
$${3+25-1\choose 25}-{3 \choose 1}{3+9-1 \choose 9}$$


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*I understand why ${3+25-1\choose 25}$ and ${3+9-1 \choose 9} $ are there, but I don't understand why ${3 \choose 1}$ is there.

*I don't understand why $x+y+z=25-16$ and not $x+y+z=25-15$.

The number of solutions to $x + y + z = 25 - 15$ is also the number of ways
to put the balls in the boxes if there must be at least $15$ balls in the first box.
But some of those solutions correspond to arrangements with only $15$ balls
in the red box, which we do not want to exclude from our final answer.
We want to subtract only the arrangements that have at least $16$ balls in
one of the boxes, so we count the number of solutions of $x + y + z = 25 - 16$.
The $\binom 31$ comes into the formula because the box that contains $16$ (or more) balls
could be the first box, the second box, or the third box.
(There can only be one such box, since $2\cdot16 > 25$.)
Each of those cases includes a number of arrangements equal to the number
of solutions of $x+y+z=25−16$.
