How to explain this 4! in the answer of this discrete math problem?

Question

The question is:

Find the number of ways to arrange letters in "bookkeeper" and all k's must precede all of o's.

My thought: Since there are ten letters, the permutation of all letters is 10! and letter e is repeated three times, so we need to divide 3! to exclude duplicate cases.

The right answer is $\frac{10!}{4!3!}$. But I do not know how to explain 4!.

My professor told me that I could think two k's as two o's so there four o's. But I still not truly understand the logic behind that.

What if the question is put all e's precede all o's?Can we still think three e's as o's so there are five o's?

Also, any resources for practice this kind of questions?

Answer 1

What your professor is saying is, let's consider Ks and Os as Xs. Then there are $\frac{10!}{4!3!}$ ways to arrange the string.

In the generated strings, since both Ks must be before any of the Os, the first two Xs must be Ks and the last 2 Xs must be Os. This gives you exactly 1 correct string with Ks and Os from the string with Xs, so their numbers are identical.

Answer 2

Think of arranging just two $k$'s and two $o$'s ... but the $k$'s have to be in front of the $o$'s ... obviously there is only one arrangement: $kkoo$ ... which is the same number of ways in which you can arrange four $o$'s. That is, since you don't have any alternatives in intermixing the $k$ with the $o$'s, we can treat them as the same.