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?

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 $\ne 0$.
