Arranging 11 letters in a row so that every Y lies between two X's (not necessarily adjacent)

Question

In how many ways can the 11 letters ABCDEFXXXYY be arranged in a row so that every Y lies between two X's (not necessarily adjacent)?

The answer is $3\binom{11}56!$ – my working: $$\begin{align} ... X Y X Y X ... \end{align}$$ so there will be 6 gaps between the X's and Y's to insert the remaining distinct letters. This is where I am stuck. I know that there are $6!$ ways to arrange the remaining 6 distinct letters, but how did the $3$ and $\binom{11}5$ come about? Is my idea/approach wrong?

Answer 1

$\binom{11}{5}$ is the number of ways to select $5$ positions among $11$. In the five selected positions, we put the letters $\{X,X,X,Y,Y\}$ in one if these $3$ possible ordered arrangements: "XXYYX", "XYYXX" and "XYXYX" (every $Y$ lies between two $X$'s). In the remaining $6$ positions, we put the 6 distinct letters $\{A, B, C,D, E, F\}$ in $6!$ ways. Therefore the total number is $$\binom{11}{5}\cdot 3\cdot 6!.$$

Answer 2

There are $11\choose 5$ ways to pick the positions of the $X$s and $Y$s (without deciding which of them is $X$ and which is $Y$). The sequence of the $X$s and $Y$s alone must begin with $X$ and end with $X$. This leaves $3\choose 1$ ways to arrange the remaining three letters (one $X$ and two $Y$). Finally, there are $6!$ ways to arrange the six different letters $A,\ldots, F$ in the remaining six positions.