How much 3x4 matrix with integer numbers (every num >= 0) , which in every row the sum is 3 , and there no column of zero's? I tried to answer the question and get a result of 7998 options , that is right?
from my way of solution , there are 20^3  3x4 matrixes which sum of every row is 3, and then i tried to substract num of options of zero's columns , according to  inclusion-exclusion , i got only 4-6+4-0 = 2 options , and then 20^3 - 2 = 7998.
 A: There are obviously far more than two matrices that don't work because of the no-zero-columns restrictions:  For one thing, each individual layout for a single row, if tripled up, will have a zero column in it, and that's 20 in itself.
There are $3 + 6 + 1 = 10$ possible rows that have (at least) a zero in a specified column - that is, there are 10 rows that have zeroes in the first column, and 10 rows that have zeroes in the second column, etc; similarly there are $2 + 2 = 4$ possible rows that have a zero in two specified columns, and $1$ possible row that has a zero in three specified columns.
In order to not be included in our list of valid matrices, there must be a specified column that is all zeroes.  For each specified column there are thus $10^3$ invalid matrices, and since there are $\binom41=4$ ways to specify a single column, we must subtract $4\cdot10^3$ entries.  But now we've taken out too many: if a matrix has two empty columns, then we've removed it twice.  We're in inclusion-exclusion land!
There are $4^3$ invalid matrices for each possible pair of specified columns and $\binom42=6$ ways specify two columns, so $6\cdot4^3$, and $1^3$ invalid matrices for each possible trio of specified columns and $\binom43=4$ ways to pick such a trio, so $4\cdot1^3$.
So our final score is
$$20^3-\binom41\cdot10^3+\binom42\cdot4^3-\binom43\cdot1^3=4\,380$$
valid matrices.
