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?

  • $\begingroup$ 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

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 $3!$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.