Question 252789: There are five balls that must be played in a game. Two of the balls are signed
balls and must be played either first or last. Each ball can only be played once.
How many possible orderings (permutations) can the balls be played?
A) 2 B) 12 C) 60 D) 120 E) 3125
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! each signed ball have to be played in the first or last game so they are fixed at 2 games, one for each.
the other 3 balls can be in games 2,3,4 in any order.
all balls can be played in one game only.
the number of permutations for the 3 unsigned balls is 3! = 3*2*1 = 6.
the two signed balls can be played in the first or last position only.
the number of permutations for the signed balls is 2!.
the total order of permutations is 2 * 6 = 12.
that's 2 permutations for the signed balls times 6 permutations for the unsigned balls.
here's how it works.
let a,e be the signed balls.
let b,c,d be the unsigned balls.
your possible permutations for games 1,2,3,4,5 are:
a,b,c,d,e
a,b,d,c,e
a,c,b,d,e
a,c,d,b,e
a,d,b,c,e
a,d,c,b,e
e,b,c,d,a
e,b,d,c,a
e,c,b,d,a
e,c,d,b,a
e,d,b,c,a
e,d,c,b,a
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