*Posted by Sreenath in : potd , trackback
*
Albert, Billy, Mitchell, Seungln, and Saketh are standing in a line. They rearrange themselves randomly such that Seungln is not next to Albert or Saketh. There is a spider on the ceiling above the first spot in the line. What is the probability that the spider is directly above Seungln?

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First, consider the number of arrangements in which Seungln is directly under the spider. Then Billy or Mitchell can be second, and there are $3!$ ways to order the other three. $2\cdot3!=12$

The number of ways Seungln can be on the opposite edge (farthest from the spider) is also $12$ by symmetry.

For the three spots in the middle, Billy and Mitchell are both next to Seungln, and Saketh and Albert are not. There are two ways to arrange each of these pairs, and thus $2\cdot2=4$ arrangements overall.

The answer is then $$\frac{12}{12+4+4+4+12}=\boxed{\frac{1}{3}}$$

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