Question 252789
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