How many integer solutions to the equation (discrete math) How many integer solutions to the equation $x_1 + x_2 + x_3 = 15$ so that $x_1 \ge 3, x_2 \ge 2$ and $x_3 \ge 0$ ? I honestly don't know where to start on this
Thanks
 A: HINT: Let $y_1=x_1-3$, $y_2=x_2-2$, and $y_3=x_3$. Note that $x_1+x_2+x_3=15$ if and only if $y_1+y_2+y_3=10$, and integers $x_1,x_2$, and $x_3$ satisfy the inequalities $x_1\ge 3$, $x_2\ge 2$, and $x_3\ge 0$ if and only if the corresponding $y_1,y_2$ and $y_3$ are non-negative. Now use the standard stars-and-bars solution to the problem of counting solutions to $y_1+y_2+y_3=10$ in non-negative integers.
A: Let us consider a set |S| which contains all the integers whose sum is 15 and which satisfies all the condition. Continuing from the part where Brian M. Scott has solved it beautifully sum y(i) = 10 bt considering the orignal Question.. 
continuing in order to get a sum of 15 u will be requiring 15 values for one x1 bt if there are three values then the options also increases ... 
example if x1 + x2 = 4 then the {(0,4) (4,0) (1,3) (2,2) (3,1) } will be the solution the number of sets are 5 or 5+2-1 C 5.. Now considering the set A in which the values are negative then compliment of A will give us the answer so solving 
A1' intersection A2' iterrsection A3' = |S| - |A1| - |A2| - |A3| + |A1 intersection A2| + |A1intersection A2 intersection A3 |....
calculation of A1 =   15 - 3 +3 -1 C 12 the first 3 has been substracted because the value has to be gr8er than 3 the second three is added because 3 is the number of variables.
Similarly A2 and A3 while calculating intersection we will be substracting both of the limits. 
A1 intersection A2 = 15 -3 -2 + 3-1 C 10..
A: Let's consider the number of solutions of $x_1 + x_2 + x_3 = n$, where $x_1 \in \{3, 4, ...\}$, $x_2 \in \{2, 3, ...\}$ and $x_3 \in \mathbb{N} \cup \{0\}$.
Let $a_n$ be the number of solutions of that equation. You're looking for $a_{15}$.
We can easily find the generating function of $a_n$:
$$g(x) = (x^3 + x^4 + ...)(x^2 + x^3 + ...)(1+x+x^2+...)$$
So that the coefficient of $x^n$ is $a_n$:
$$g(x) = \displaystyle{\sum_{n=0}^\infty a_n x^n}$$
Therefore, for $|x|<1$, we conclude that:
$$g(x) = \dfrac{x^3}{1-x} \cdot \dfrac{x^2}{1-x} \cdot \dfrac{1}{1-x} = x^5 \cdot (1-x)^{-3}$$
Using the Generalized Binomial Theorem, and since $\binom{-\alpha}{k} = \binom{\alpha + k - 1}{k}(-1)^k$:
$$g(x) = x^5 \displaystyle{\sum_{k=0}^\infty \binom{-3}{k} (-1)^k x^k} = x^5 \displaystyle{\sum_{k=0}^\infty \binom{2+k}{k}x^5} = \displaystyle{\sum_{k=0}^\infty \dfrac{(k+1)(k+2)}{2}x^{k+5}}$$
Therefore:
$$
a_n = \left \{
      \begin{matrix} 
         0 & n < 5 \\
         \dfrac{(k+1)(k+2)}{2} & n \geq 5      \end{matrix}
   \right .
$$
