I need to find how many solutions of the diophantine equation in the title of the question exist with two of the following conditions and am having major trouble:
I used the repetitive combinations formula for the previous questions, but am stumped on how to precede here.
On
I think part 2 is simpler here.
Your answers will be of the form: $(0, 2, 2, 4, 2, 4)$ where only one spot is 0 and only two spots are 4 or $(0, 2, 2, 2, 2, 6)$ where only one spot is 0 and only one spot is 6.
How many of these 6-tuples can you generate? For ex. Of the second kind you gotta pick a place for the 0 and a place for the 6, resulting in 6*5 = 30 possible 6-tuples.
For question 1, just go by:
How many different forms can you have? Some examples $(0, 0, 0, 0, 0, 14), (0, 6, 6, 2, 0, 0), (2, 2, 4, 4, 2, 0)$
It can get really boring to count them this way, so you can generate simpler formats and use other shorter ways to count the solutions. However, that is more advanced and this looks more like homework to me
For one: Let them all be even first (so that $x_i=2k_i$ for some integers $k_i$). Then, you can rewrite: $$x_1+x_2+\ldots+ x_6=14\implies k_1+k_2+\ldots+k_6=7$$ and that's your typical partition problem (usually done with the "bars and stars" method). For the second one, first let one of the variables equal $0$ (say $x_1=0$). Then, how many ways can you have $$x_2+x_3+\ldots +x_6=14$$ is again your normal partition problem. Do this for each $x_i$. (Notice any symmetries to make calculations a little easier for this one?)