Glass Balls

Label the balls in the first column with an "A", in the second column with a "B", in the third column with a "C" and in the fourth column with a "D".

Then the number of different ways in which the $16$ balls can be broken is just the number of ways in which $4$ A's, $4$ B's, $4$ C's and $4$ D's can be arranged in a row. So the answer is

$\dfrac{16!}{4!*4!*4!*4!} = \boxed{63063000}$ .

Note that, since it is always the lowest ball in any column that gets broken, when we come to any letter in the row of A's, B's, C's and D's it is automatically known which ball is being struck. Thus we didn't have to distinguish between balls in any given column, as the sequence in which they are shot, relative to one another, is a given.

Let's say I give the marksmen an ordered arrangement of 16 numbers of which 4 are 1s, 4 are 2s, 4 are 3s and 4 are 4s. Now I instruct him to start with the 1st number in that arrangement and shoot the lowermost ball of the string corresponding to that number and keep doing that in the order of numbers I gave him.

For example, let's say, I gave him:

1 2 4 3 3 4 1 3 2 1 4 2 1 2 3 4.

Now, he shoots the lowermost ball of 1st string, then of second string, then of 4th string, then of 3rd string, then again of 3 string and so on.

We can easily see that the number of ways in which I can give him the instruction is the required answer.

Number of such possible arrangements can be given by

$\frac { 16! }{ 4!\times 4!\times 4!\times 4!\quad } =\quad 63063000$

This is the same question as asking how many ways can you arrange AAAABBBBCCCCDDDD.

16!/4!4!4!4! = 63063000