Combination Problem on white table

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?

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

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