constructive counting math problem about checkers on a checkerboard In how many ways can we place anywhere from 0 to 9 indistinguishable checkers on a 3x3 checkerboard (no more than one checker per square), such that no row or column contains exactly 1 checker?
I tried reducing the checkerboard to a 2x2 checkerboard and solving it that way, but it didn't work out and now I'm not sure what to do/how to re-start. 
 A: You can of course use $0$ checkers to get one arrangement. It’s not hard to see that there are no acceptable arrangements of $1,2$, or $3$ checkers, and that the acceptable arrangements of $4$ checkers are precisely those in which they occupy the corners of a rectangle with sides parallel to the sides of the board; I’ll leave it to you to count those.
Once you get up to $5$ checkers, it’s easier to look at the unoccupied squares, since there are at most $4$ of them. The condition that no row or column contain exactly one checker is equivalent to the condition that no row or column contain exactly two empty squares. Thus, we need to determine in how many ways we can arrange up to $4$ empty squares so that no row or column contains exactly $2$ of them. Clearly there is one way with $0$ empty squares, and there are $9$ ways with $1$ empty square. Two empty squares need to be in different rows and columns; pick a location for one of them, and there are exactly $4$ possible locations for the other, for a total of $9\cdot4=36$ ordered pairs of locations. However, these counts each unordered pair twice, so we get only $18$ actual arrangements.
Three empty squares must either occupy a row or column or have one in each row and column; counting those arrangements isn’t too hard, and I’ll leave it to you. I’ll also leave the case of $4$ empty squares to you; you may find the pigeonhole principle useful in thinking about what can happen.
A: We can also do this using inclusion-exclusion. There are $\binom3j\binom3k$ ways to choose $j$ particular rows and $k$ particular columns for which the restriction is violated. Due to the symmetry with respect to $j$ and $k$, we have $4(4+1)/2=10$ cases to consider:
$j=0$, $k=0$, $2^9=512$ arrangements
$j=0$, $k=1$: $3\cdot2^6=192$ arrangements
$j=0$, $k=2$: $3^2\cdot2^3=72$ arrangements
$j=0$, $k=3$: $3^3=27$ arrangements
$j=1$, $k=1$: $(1+2\cdot2)\cdot2^4=80$ arrangements
$j=1$, $k=2$: $(2\cdot2+2\cdot2)\cdot2^2=32$ arrangements
$j=1$, $k=3$: $3\cdot2\cdot2=12$ arrangements
$j=2$, $k=2$: $(1+2+4)\cdot2^1=14$ arrangements
$j=2$; $k=3$: $3!=6$ arrangements
$j=3$, $k=3$: $3!=6$ arrangements
Thus by inclusion-exclusion there are
$$
\binom30\binom30\cdot512-2\cdot\binom30\binom31\cdot192+2\cdot\binom30\binom32\cdot72-2\cdot\binom30\binom33\cdot27+\binom31\binom31\cdot80-2\cdot\binom31\binom32\cdot32+2\cdot\binom31\binom33\cdot12+\binom32\binom32\cdot14-2\cdot\binom32\binom33\cdot6+\binom33\binom33\cdot6\\=512-1152+432-54+720-576+72+126-36+6\\=50$$
admissible arrangements.
