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How many ways can we place $3$ balls in $4×4$ grid such that no $2$ balls in the same row or column?

My try follows

Choose $3$ rows out of $4$ by $4C3= 4$

first ball has $4$ choices

Second ball has $3$ choices

Third ball has $2$ choices

So total number = $4×4×3×2=96$

Is my work right?

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    $\begingroup$ Yes, that is fine. You may also wish to look up rook polynomials. $\endgroup$ – Lord Shark the Unknown Aug 23 '17 at 20:06
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    $\begingroup$ Another way to figure is the first ball can be put in $16$ places, the second ball $9$ and the third ball in $4$ places ... & each configuartion can be formed in $3!$ possible ways ... so $16 \times 9 \times 4/6$ which is indeed $\color{red}{96}$. $\endgroup$ – Donald Splutterwit Aug 23 '17 at 20:09
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    $\begingroup$ Yes, if the balls are indistinguishable, which you didn't tell us. $\endgroup$ – bof Aug 24 '17 at 10:33
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You can choose a row and a column as undesirable in $4\cdot 4=16$ ways. Then you can select the three balls from the remaining array in as many ways as a $3\times3$ determinant has terms, namely $6$. Hence there are 96 admissible selections in total.

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Yes, the answer is $$\binom{4}{3}4!=96$$ for precisely the reasons you describe (if the balls are indistinguishable).

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