Finding the number of solutions of the equation $c_1 + c_2 + c_3 = 20$ in the nonnegative integers if $c_2$ is even and $c_3$ is odd Find the number of nonnegative integer solutions to the equation $c_1 + c_2 + c_3 = 20$, where $c_2$ is even and $c_3$ is odd.
I started with finding the total number of solutions, $C(22,20)=231$. Then I tried to apply the inclusion-exclusion principle. Let $P_1$ be the number of solutions that do not satisfy the condition for $c_1$, $P_2$ for $c_2$ and so on.
$N-P_1-P_2-P_3+P_1P_2+P_2P_3+P_1P_3-P_1P_2P_3$ gives us the desired number of solutions. $P_1, P_1P_2, P_1P_3$ and $P_1P_2P_3$ are all zero because there is no condition for $P_1$. To find $P_2$, I assigned odd numbers to the $c_2$ and then calculated the number of solutions which is $110$. $P_3$ is also $110$. If I'm right up to this point, I'm stuck at finding $P_2P_3$.
To find the number of solutions that do not satisfy $c_2$ and $c_3$, I assigned $c_2$ odd numbers and $c_3$ even numbers. Adding an even and an odd number gives us an odd number. From what I previously calculated, I think there is a total of $110$ solutions again.
Thus, $231-110-110+110=121$. Is this true?
For reasons I will explain below, it is easier to solve this problem directly than to use the Inclusion-Exclusion Principle.
A direct approach:
Since $20$ is an even number, $c_2$ is odd, and $c_3$ is even, $c_1 = 20 - c_2 - c_3$ is also an odd integer.  Thus, there exist nonnegative integers $x_1, x_2, x_3$ such that $c_1 = 2x_1 + 1$, $c_2 = 2x_2$, and $c_3 = 2x_3 + 1$.
\begin{align*}
c_1 + c_2 + c_3 &amp; = 20 \tag{1}\\
2x_1 + 1 + 2x_2 + 2x_3 + 1 &amp; = 20\\
2x_1 + 2x_2 + 2x_3 &amp; = 18\\
x_1 + x_2 + x_3 &amp; = 9 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers which has the same number of solutions as equation 1 subject to the stated constraints.  A particular solution of equation 2 corresponds to the placement of two addition signs in a row of nine ones.  For instance,
$$++ 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $x_1 = x_2 = 0$, $x_3 = 9$ (and $c_1 = 1, c_2 = 0, c_3 = 19$), while
$$1 1 1 1 + 1 1 1 + 1 1$$
corresponds to the solution $x_1 = 4$, $x_2 = 3$, $x_3 = 2$ (and $c_1 = 9, c_2 = 6, c_3 = 5$).  The number of solutions to Equation 2 is
$$\binom{9 + 3 - 1}{3 - 1} = \binom{11}{2}$$
since we must select which two of the eleven positions required for nine ones and two addition signs will be filled by addition signs.
Remarks about your approach:
If you define $P_i$ to be the number of solutions that violate the constraint on $c_i$, then $P_iP_j$ is the product of $P_i$ and $P_j$.  Presumably, you meant $P_iP_j$ is the intersection of the sets of solutions that violate the constraint on $c_i$ and the constraint on $c_j$.
If you wish to the Inclusion-Exclusion Principle, you should define $A_i$ to be the set of solutions that violate the constraint on $c_i$, in which case the number of admissible solutions is
$$N - |A_1 \cup A_2 \cup A_3|$$
where
\begin{align*}
|A_1 \cup A_2 \cup A_3| &amp; = |A_1| + |A_2| + |A_3|\\
                        &amp; \quad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3|\\ 
                        &amp; \qquad + |A_1 \cap A_2 \cap A_3|
\end{align*}
As stated above, the constraint that $c_2$ is even and $c_3$ is odd imposes the additional constraint that $c_1$ is odd.
Let's examine $|A_1|$.  If the condition that $c_1$ is odd is violated, then we need to find the number of solutions of the equation
$$c_1 + c_2 + c_3 = 20$$
subject to the constraint that $c_1$ is even.  If $c_1$ is even, then either $c_2$ and $c_3$ are both even or they are both odd.  If $c_2$ and $c_3$ are both odd, we need to find the number of solutions of the equation $c_1 + c_2 + c_3 = 20$ in which exactly two of the variables are odd, that is, we need to solve the original problem in order to use the Inclusion-Exclusion Principle. Therefore, it makes more sense to solve the problem directly.
Alternative approach:
Any solution to $c_1 + c_2 + c_3 = 20$ where $c_2, c_3$ are of opposite parity (i.e. one odd, the other even) will occur if and only if $c_1$ is odd.  Further, for each of these solutions, you know by symmetry, that exactly $(1/2)$ of them will represent $c_2$ even and $c_3$ odd.
So, you may assume that $c_1 \in \{1,3,5,\cdots, 19\}.$
Further, for each such value of $c_1$, the problem will be represented by :
$c_2 + c_3 = (20 - c_1)$, which will have
$\displaystyle \binom{[20 - c_1] + [2-1]}{[2-1]} = (21 - c_1)$ solutions.
Therefore, the desired tally is
$$\left(\frac{1}{2}\right) \times \sum_{k = 1}^{10} (21 - [2k-1]) ~=~ \left(\frac{1}{2}\right) \times \sum_{k = 1}^{10} (22 - [2k]).$$
The right hand factor above is $220 - 2(55) = 110$.
Therefore, the final computation is $(110/2) = 55.$