# how many ways can targets be broken if only the target on a bottom of a row can be broken?

Twelve clay targets (identical in shape) are arranged in four hanging columns. There are four red targets in the first column, three white ones in the second column, two green targets in the third column and three blue ones in the fourth column. To join her college drill team, Deborah must break all 12 of these targets (using her pistol and only 12 bullets) and in so doing must always break existing target at the bottom of a column. Under these conditions, in how many different orders can Deborah shoot down (and break) the 12 targets?

The answer key gives $\frac{12!}{4!3!2!3!}$ but I don’t see why? She definitely has 4 options to choose from the first shot, 4 options to choose from the second shot, but then she could have broken the two in the third column and I’m not sure where to go from here. I don’t see how this question is a matter of counting permutations and removing repetitions?

There are $12!$ different sequences of targets. The restriction that she can only shoot the bottom target means that she has only one option of each color. Thus the number of ways she can shoot the targets is the number of possible sequences of $4$ red, $3$ white, $2$ green and $3$ blue targets, where targets of the same color are indistinguishable. Thus the number of sequences of targets she could shoot is $$\frac{\text{# of sequences of targets}}{\text{# of sequences of red}\times\cdots\times \text{# of sequences of blue}}=\frac{12!}{4!\times 3!\times 2!\times 4!}$$
The sequence used by Deborah is completely described by a word of length $12$ obtained by arranging the letters in the word RRRRWWWGGBBB.