1
$\begingroup$

What is the total possible no. of ways that $4$ shoes can be chosen from $5$ pairs of shoes such that none of the shoes chosen forms a pair ?

So say the shoes are $(A_1,A_2); (B_1,B_2); (C_1,C_2);(D_1,D_2);(E_1,E_2)$. I have to choose $4$ among the $A,B,C,D,E$ s such that no same letter appears. So one from each pair, so $2$ choices for each draw, giving $2^4$ choices, but this is multiplied with how many ways $4$ letters can be chosen from $5$ letters. Hence ${5\choose 4} \times2^4$ choices.

Is this correct ?

Please help

$\endgroup$
  • $\begingroup$ Your solution is correct. However, this is not a derangement problem. A derangement is a permutation that leaves no object in its original position. $\endgroup$ – N. F. Taussig Sep 3 '18 at 22:26
  • $\begingroup$ @N.F.Taussig: oh I see ... I didn't know that ... I just cooked up a name ... may be non-pairing would be more appropriate ? $\endgroup$ – user521337 Sep 3 '18 at 23:04
0
$\begingroup$

Yes that is perfectly correct answer and procedure.

$\endgroup$

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.