# number of solutions for this equation

What is the number of solutions for this equation: $$x_1 + x_2 + x_3 = 15$$ when $$x_1 \in \mathbb{N}_{even} \cup \{0\}$$, $$0\leq x_2\leq 5$$, $$x_3 \geq 0$$?

Can it be solved without generating functions? If not, how can I use generating functions to solve it?

EDIT: $$x_1, x_2, x_3$$ can be only integers

• There are so few possibilities you can just list them. – saulspatz Jan 12 '19 at 17:53
• @saulspatz I have counted at least 20, it seems like a lot of work – Robo Yonuomaro Jan 12 '19 at 17:58
• Can $x_2,x_3$ be real or just integers are alowed? – Patricio Jan 12 '19 at 18:09
• Integers, I forgot to mention it – Robo Yonuomaro Jan 12 '19 at 18:10

Start by putting $$x_1=2k$$ (where $$k$$ varies from $$0$$ to $$7$$). This gives us the following

$$x_2 + x_3 = 15-2k = n$$

Now, using the stars and bars rule we know that the above equation has following number of integral solutions for any particular $$k$$

$$N_1 = \binom{n+2-1}{2-1} = \binom{n+1}{1} = \binom{16+2k}{1}$$

Now from this we subtract all the cases in which $$x_2\ge6$$. So put $$x_2=y_2+6$$ which gives us $$y_2\ge0$$

$$y_2 + x_3 = 15-2k-6 = 9-2k$$

Number of integral solutions to this are

$$N_2 = \binom{9-2k+2-1}{2-1} = \binom{10-2k}{1}$$

Summing up $$N_1$$ from $$k=0$$ to $$k=7$$

$$\sum_{k=0}^7\binom{16+2k}{1} =\sum_{k=0}^716+2k=184$$

Now summing up $$N_2$$ from $$k=0$$ to $$k=4$$ as after that the terms will simply be $$0$$ in the binomial coefficient

$$\sum_{k=0}^4\binom{10-2k}{1} =\sum_{k=0}^410-2k = 30$$

Subtracting $$N_2$$ from $$N_1$$, we'd get

$$N=N_1-N_2 = 184-30 =154$$ which is the total number of possibilities.