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My solution is:

  1. Choose $1$ man from $6$ men: $6$ ways
  2. Choose $1$ woman from $6$ women: $6$ ways
  3. Arrange the rest of the people: $10!$ ways
  4. Insert the pair in the arranged line: $11$ ways
  5. Swap a man and a woman in the pair: $2$ ways

Therefore, the result is $6 \cdot 6 \cdot 10! \cdot 11 \cdot 2 = 11! \cdot 72$

But my friends argue that we don't have to choose a man and a woman in the pair because the problem already told us that there is just $1$ man and $1$ woman that must stand next to each other. My argument is that if we don't choose which pair it is, we can't write the outcomes because we don't know which pair it is.

Which one is the right way?

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  • $\begingroup$ Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Your friends have interpreted the question correctly. We are told that a particular man and a particular woman stand next to each other, so you should divide your answer by $36$. $\endgroup$ – N. F. Taussig Jul 11 '18 at 0:26
  • $\begingroup$ Can write the outcome with that interpretation? $\endgroup$ – troy_KD Jul 11 '18 at 0:31
  • $\begingroup$ @N.F.Taussing Can you explain what the OP wants? $\endgroup$ – stressed out Jul 11 '18 at 0:31
  • $\begingroup$ I would think of the problem as "How many ways are there for 6 men and 6 women to stand in a line if Fred and Ginger must stand together?" (Personally, I find it easier to visualize the situation if I give the people names.) $\endgroup$ – awkward Jul 11 '18 at 12:23
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Your friends have interpreted the question correctly. The particular man and woman who stand next to each other are given. Therefore, the first two steps of your proposed solution are superfluous. The final three steps, which you did correctly, yield the answer $10! \cdot 11 \cdot 2 = 2 \cdot 11!$.

Alternate Solution: Treat the particular man and woman who must stand next to each other as a single object. Then we have eleven objects to arrange, a block containing the man and woman in question and the other ten people. The eleven objects can be arranged in $11!$ ways. The man and woman can be arranged within the block in $2!$ ways. Hence, there are $11!2!$ admissible arrangements.

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