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Question : How many ways are there to choose 3 cells from a 4x4 table such that any two chosen cells do not belong to the same row nor the same column?

What have i done so far :

choosing $1$ from $16$ cells, each giving remaining $9$ cells option, so my equation would be $$(16)(9)(4)=576$$ which my answer was obviously false since $16C3=560$

Please explain where did I go wrong? What shall be the correct approach to this answer?

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  • $\begingroup$ Hint: How many ways are there to choose a given set of three cells that satisfy the constraint? $\endgroup$ – Yly Jan 27 at 4:48
  • $\begingroup$ Also, 16C3 is irrelevant here, since it does not take the constraint of no common rows or columns into consideration, and hence overcounts. $\endgroup$ – Yly Jan 27 at 4:49
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I think there are some double countings in your method.

In the same shape, the method includes $3!$ counts, so my opinion is to divide it into $3!$.

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