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. 

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.
