Four people line up for a roller coaster.

combinatorics/roller.coaster New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

Four people line up for a roller coaster.  One of them refuses to be
first in line.  In how many ways is it possible for the people to line
up?

-- joseph mccann <jomac@globalnet.co.uk>

Show solution

Assume person #1 is the coward.  That leaves 3 positions that he is
willing to line up in.  For each position of person #1 there are 3!
combinations for the other 3 people to line in.  Hence:  3(3!) = 18.

-- Craig Jones <CMJones@Microsoft.com>