Through a marksmanship contest, there are 4 strings of 4 glass balls hanging down from a horizontal post. Each bullet can only hit one glass ball at a time, so 16 shots need to be fired.

The only problem is if you shoot a glass ball that has a glass ball hanging below it (on the same string), it will fall off. So given the rule that you can't shoot a glass ball with a glass ball underneath it (and on the same string), how many ways can you shoot all the glass balls?

This answer is the same as Charlie's, but not half as elegant as his.

If the balls are hanging in this formation:

b01....b02....b03....b04
b05....b06....b07....b08
b09....b10....b11....b12
b13....b14....b15....b16

Then the order of balls shot must include these sequences:

{b13, b09, b05, b01}
[b14, b10, b06, b02}
{b15, b11, b07, b03}
{b16, b12, b08, b04}

If N is all the ways to merge these four sequences that preserve the internal sequence orders, the N*[(4!)^4)] = all the ways to order 16 balls = 16!

N = (16!)/[(4!)^4]
= (20,922,789,888,000)/(331,776)

N=63,063,000 = the number of ways to shoot the glass balls

Posted by Penny on 2004-01-31 05:02:27