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!
Best Answer
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 ${21 \choose 2} = 210$.