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

  • $\begingroup$ 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

Yes that is perfectly correct answer and procedure.


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