<s>[INST] 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.
 [/INST] $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}$
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