Number of possibilities to rearrange a word such that a character is to the right of the other

Question

The question:

In how many ways can we arrange the word $FOOTBALL$ in a row such that $A$ will appear to the right of $F$ (but not necessarily coupled).

If I scale down the problem I can see that this condition divides the possibilities in 2. Therefore having $$\frac{8!}{2!*2!*2}$$

Why does this condition divides the number of ways in 2?

Answer 1

As you said, there are $\frac{8!}{2! \cdot 2!}$ ways of permutation. Because there is only one instance of $A$ and one instance of $F$, and no two character can be placed in the same position, this leads us to say that because $A$ is either to the left or to the right of $F$ (you could say that vice versa, too) then in exactly half of these permutations the letter $A$ is to the left of the letter $F$, and in the other half the letter $A$ is to the right of letter $F$.

From here all you need to do is divide the total number of permutations by $2$ to get your answer.

Answer 2

Because if $X$ is a valid arrangement, then swapping F and A in $X$ gives an invalid arrangement, and vice versa.

Answer 3

The given answers already cover your question. But, if you want to solve this problem in a different manner, probably because you find the current answer confusing, here it is:

Fix the two places $F$ and $A$ will occupy in $^8C_2$ ways (since $A$ can only occur to the right of $F$). The remaining letters can be permuted in the remaining places in $\frac{6!}{2!2!}$ ways.

So, total ways are $= ^8C_2\frac{6!}{2!2!}=\frac{8!}{6!2!}\frac{6!}{2!2!}=\frac{8!}{2!2!\cdot2}$ .