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?
 A: 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.
A: The number of nonnegative integer solutions of $x_1 + x_2 + x_3 = 11$ is the coefficient of $t^{11}$ in the following generating function [JDL]
$$\dfrac{1}{(1-t)^3}$$
Suppose now that we are interested in integer solutions with $x_1 \geq 1$, $x_2 \geq 2$ and $x_3 \geq 3$. We thus introduce three new nonnegative variables
$$z_1 : = x_1 - 1 \qquad\qquad\qquad z_2 : = x_2 - 2 \qquad\qquad\qquad z_3 : = x_2 - 3$$
The number of admissible integer solutions of $x_1 + x_2 + x_3 = 11$ is the number of nonnegative integer solutions of $z_1 + z_2 + z_3 = 5$, which is the coefficient of $t^5$ in the generating function.
Using SymPy:
>>> from sympy import *
>>> t = Symbol('t')
>>> f = 1 / (1-t)**3
>>> f.series(t,0,12)
1 + 3*t + 6*t**2 + 10*t**3 + 15*t**4 + 21*t**5 + 28*t**6 + 36*t**7 + 45*t**8 + 55*t**9 + 66*t**10 + 78*t**11 + O(t**12)

Hence, the number of admissible integer solutions is $21$. Note that the coefficients in the series are the triangular numbers (A000217)
$$\binom{2}{2}, \binom{3}{2}, \binom{4}{2}, \binom{5}{2}, \dots, \binom{k+2}{2}, \dots$$

[JDL] Jesús A. De Loera, The Many Aspects of Counting Lattice Points in Polytopes.
A: $$(1,0,10),(2,3,6),(1,2,8),(1,3,7)$$ which satisfies above equation, if rearranged gives 24 solutions and you have even more solutions. So your text book answer is wrong and I dont see any problems in ur method.
A: This can be solved also using the stars and bars method.
The point is paying attention to variables that take value 0.
So you have 3 cases:
1) all variables $\ne 0$
This amounts to $\binom{11-1}{3-1}=45$
2) just one variable have value $0$ (and hence two others are $\ne 0$)
This amounts to $\binom{3}{1}\cdot\binom{11-1}{2-1}=30$
3) two variables are $0$ (and hence only one is non-zero
This amounts to $\binom{3}{2}\cdot 1=3$
Taking 1)+2)+3) gives 78
as already found with the other methods.
For the second part, you just have to adjust your question to the new constraints, that is to say $x_1+x_2+x_3=8$ (can you see this?). Applying again the stars and bars method you find
$\binom{8-1}{3-1}=21$
The situation is simpler in this second part since all variables are $\ne 0$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
\bracks{z^{11}}\sum_{x_{1} = 1}^{\infty}z^{x_{1}}
\sum_{x_{2} = 2}^{\infty}z^{x_{2}}\sum_{x_{3} = 3}^{\infty}z^{x_{3}} & =
\bracks{z^{11}}{z \over 1 - z}\,{z^{2} \over 1 - z}\,{z^{3} \over 1 - z} =
\bracks{z^{\color{#f00}{5}}}\pars{1 - z}^{-3}
\\[5mm] & = \bracks{z^{5}}\sum_{i = 0}^{\infty}{-3 \choose i}\pars{-z}^{i} =
-{-3 \choose 5} = {7 \choose 5} = {7 \times 6 \over 2} = \bbx{\ds{21}}
\end{align}

Note that $\ds{\color{#f00}{5} = 11 - 1 - 2 - 3}$.

