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I actually got this part and I got $16$ choose $2$, which would be $120$. The part I didn't get which wouldn't fit into the title was in how many of these solutions is $x_1 \geq 1$, $x_2 \geq 2$,and $x_3 \geq 3$? I'm not very good at combinatorics so I don't really know. I think the way to approach it would be saying how many solutions where $x < 1 +$ solutions where $x < 2 +$ solutions where $x < 3$ and subtract that from $120$, but I don't know how to do that.

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Your solution to determining how many solutions the equation $$x_1 + x_2 + x_3 = 14 \tag{1}$$ has in the nonnegative integers is correct. To determine how many solutions equation 1 if $x_1 \geq 1$, $x_2 \geq 2$, and $x_3 \geq 3$, let \begin{align*} y_1 & = x_1 - 1\\ y_2 & = x_2 - 2\\ y_3 & = x_3 - 3 \end{align*} Then $y_1, y_2, y_3$ are nonnegative integers. Substituting $y_1 + 1$ for $x_1$, $y_2 + 2$ for $x_2$, and $y_3 + 3$ for $x_3$ in equation 1 yields \begin{align*} y_1 + 1 + y_2 + 2 + y_3 + 3 & = 14\\ y_1 + y_2 + y_3 & = 8 \end{align*} which is an equation in the nonnegative integers with $$\binom{8 + 2}{2} = \binom{10}{2} = 45$$ solutions.

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This is pretty easy to calculate through Wolfram Alpha: Here is an example of an input that will solve this for you. The answer provided is not entirely true though, because of your constraints. We must exclude the solution $12+1+1$, which does not have $x_2 \geq 2$. We thus subtract $1$ from the provided result of 16 to get a final answer of $\color{red}{15}$. For clarification I have provided a list of the solutions here. If you want further info check out this page, although the topic of integer partitions is fairly complicated. If you have further questions I can try to help point you in the right direction, although I am not the best source of info on this topic; if you desire to learn more I recommend creating a new question specifically geared towards this.

$$ 11 + 2 + 1 = 14\\ 10 + 3 + 1 = 14\\ 10 + 2 + 2 = 14\\ 9 + 4 + 1 = 14\\ 9 + 3 + 2 = 14\\ 8 + 5 + 1 = 14\\ 8 + 4 + 2 = 14\\ 8 + 3 + 3 = 14\\ 7 + 6 + 1 = 14\\ 7 + 5 + 2 = 14\\ 7 + 4 + 3 = 14\\ 6 + 6 + 2 = 14\\ 6 + 5 + 3 = 14\\ 6 + 4 + 4 = 14\\ 5 + 5 + 4 = 14$$

Note: In the event you also desire for the condition $x_1 \neq x_2 \neq x_3$ to hold it appears we get $10$ solutions

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