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 The glass balls (Posted on 2004-01-30)
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?

 See The Solution Submitted by Gamer Rating: 4.0000 (3 votes)

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 Solution Comment 5 of 5 |
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

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