<s>[INST] How many 3-tuples satisfy $x_{1} + x_{2} + x_{3} = 11;$ $(x_{1} ,x_{2} ,x_{3}$ are nonnegative integers?) I know that the total number of choosing without constraint is 
$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13·12}{2} =78$
Then with x1 ≥ 1, x2 ≥ 2, and x3 ≥ 3. 
the textbook has the following solution 
$\binom{3+5−1}{5}=\binom{7}{5}=21$ I can't figure out where is the 5 coming from?
The reason to choose 5 is because the constraint adds up to 6? so 11 -6 =5?
 [/INST] For the given equation: $$x_1+x_2+x_3 = 11, \;\text{ with } x_1, x_2, x_3 \;\text{ non-negative },$$  your solution is correct.
$$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13!}{2!11!} = \frac{13·12}{2} =78$$
Your final answer is correct, (Now corrected: (but you failed to show that you need to divide $13\times 12$ by $2$ to obtain $78$). 

The other solution (from the text) would be the solution to $x_1+x_2+x_3 = 5$, with $x_1, x_2, x_3$ non-negative.  In this case, there are $$\binom{3+5-1}{5} = \frac{7!}{2!5!} = \frac{7\cdot 6}{2} = 21$$

Edit after another question-update: We now are solving $$x_1 + x_2 + x_3 = 11, \text{ with }\; x_1\geq 0+1,\,x_2 \geq 0 + 2,\,x_3 \geq 0 + 3.$$
We can solve in the same manner by writing $(x_1+1)+ (x_2 + 2) + (x_3+3) = 11-1-2-3 = 5$.
Then we may simply ascribe $y = x_1 + 1, y_2 = x_2+2, y_3 = x_3 + 3$ to get $$y_1+y_2 +y_3 = 5$$
From here, the text's solution (addressed above) solves the number of the required solutions.
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