# Find number of solutions of the equation $x_{1}+x_{2}+x_{3} = 41$, where $x_{1}, x_{2} and x_{3}$ are odd and non negative integers

There are two constraints to this problem:

1) $x_{1}, x_{2} and x_{3}$ are non negative integers

2) $x_{1}, x_{2} and x_{3}$ are odd

If there had been just the first constraint (non negative integer), i would have simply used "bars and stars" method and found the answer as: 43 C 2 [41 + 2 bars] = 903

But since there is another requirement that all the non-negative integral solutions must be odd as well, even after trying different approaches, am unable to solve further. Please help me out!

Thanks a lot!

Look, according to your second constraint, $x_1$, $x_2$ and $x_3$ have the following forms:
$x_1=2y_1 + 1$, $x_2=2y_2 + 1$ and $x_3=2y_3 + 1$ with $y_1$, $y_2$, $y_3$ non-negative.
So, substituting them in the equation and simplifying we have,
$y_1 + y_2 + y_3 = 19.$
Then we can apply the “bars and stars” method and get the solution ${21 \choose 2} = 210$.

Modify the method you're referring to ...

1) Guarantee that x:s are positive by reserving one "star", and initially odd. (i.e. 3 in total)

2) Keep them odd by selecting two "stars" at a time.

I think this is enough of a clue given your description of how you'd solve the more basic problem? If not, I'll be happy to fill in the rest of the details for you.