If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$? If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?
I need to solve it using combinatorics.
I've started with calculating the number of possible solutions  without restrictions and I am stuck.
 A: The straightforward way to get the result is to compute the coefficient of $X^{12}$ in $(1+X+X^2+X^4+X^5+X^6)^4$. Though computing that coefficient basically amounts to counting solutions explicitly in a systematic manner, it is not so hard to do even by hand: $$(1+X+X^2+X^4+X^5+X^6)^2=\\1+2X+3X^2+2X^3+3X^4+4X^5+6X^6+4X^7+3X^8+2X^9+3X^{10}+2X^{11}+X^{12},$$
and from there one easily sees the required coefficient is $2\times(1^2+2^2+3^2+2^2+3^2+4^2)+6^2=122$.
I can also propose a more ad hoc method that might be slightly easier (because fewer and smaller numbers are involved) if one were not allowed to use pen and paper. Most of the solutions will involve two of $a,b,c,d$ being${}<3$ and the other two being${}>3$, the exceptions being the permutations of $(0,4,4,4)$ and of $(2,2,2,6)$, which are $8$ solutions. For the remaining solutions that do have a $2$-$2$ split, there are $\binom42=6$ ways to choose the larger positions, and after we have attributed the minimum of $4$ units to those two positions, we must still distribute the remaining $12-8=4$ units over the $4$ positions, while never putting more than $2$ of those units together. By an argument similar to the one used above, that amounts to finding the coefficient of $X^4$ in $(1+X+X^2)^4$, and from $(1+X+X^2)^2=1+2X+3X^2+2X^3+X^4$ one finds that the number is $1^2+2^2+3^2+2^2+1^2=19$. All in all $6\times19+8=122$ solutions are possible.
A: $a + b + c +d = 12 \ $ where $0 \leq a, b, c, d \leq 6$ and $a, b, c, d \ne 3$
Here is how you can solve using Principle of Inclusion-Exclusion, which is straightforward as well but I have taken more space and time to explain the steps.
A) Number of unrestricted solutions for non-negative numbers
$\displaystyle  = {15 \choose 3} = 455$
B) Number of solutions where at least one of the numbers is $3$
$ \displaystyle  = 4 {11 \choose 2} - {4 \choose 2} {7 \choose 1} + {4 \choose 1} - {4 \choose 4} = 181$
Explanation: First term is choosing one of the numbers to be $3$ and rest $3$ numbers sum to $9$ but these will overcount cases where two numbers are $3$ and hence the second term subtracts cases where two numbers are $3$ each and then other two sum to $6$. Then the last term is where three of the numbers are each $3$. Please note that if three numbers are $3$, the fourth will also be $3$. Hence in the first term we count such cases $4$ times and then subtract $6$ times in the second term. As there is only one way for three or four numbers to be $3$ each, we add $3$ so we count it one time. The other way you can write it as ${4 \choose 1} - {4 \choose 4}$ but I wanted to explain the rationale behind it.
C) Number of solutions where at least one of the numbers is greater than $6$
$ \displaystyle  = 4 {8 \choose 3} = 224$
Explanation: Now if one of the numbers is $\gt 6$, none of the other numbers can be greater than $6$. We first set one of the $4$ numbers as $7$ and then count number of ways for $4$ numbers to sum to $5$.
Now B) and C) both count solutions where one of the numbers is greater than $6$ and one of the numbers is $3$.
D) Solutions where one of the numbers is greater than $6$ and one of the numbers is $3$
$\displaystyle = {4 \choose 3} {3 \choose 1} {4 \choose 2} = 72$.
Explanation: There are $4$ ways to choose one of the numbers being $7$ and one of the numbers being $3$ and then we have $3$ numbers sum up to remaining $2$ (including the number $7$ which can also be $8$ or $9$).
So the number of desired solutions is $ = 455 - 181 - 224 + 72 = \fbox {122}$
A: Let M be an event such that at least one of a, b, c, d is equal to  . Let N be the event that either of a,b,c,d are greater than  .
Let K be the total number of solutions to the equation .
We need to find 
Now,
K=  where

Now to find M, we take M as a union of events , where
.
Using Inclusion-Exclusion

Now to find N, we take N as a union of events as well such that

where 
Using Inclusion- Exclusion

Now 
Now we just need to find 

