Arrangement of letters with one pair of letters always between two letters 
How many ways are there to arrange $11$ letters, $A,B,C,D,E,F,x,x,x,y,y$ so that every $y$ lies between two $x$ (not necessarily adjacent)$?

Attempt:I fixed the letters $ x \quad y \quad y \quad x$. Totally there are 5 gaps(all distinct) for the remaining 7 letters(these are all distinct)  So there are $5^7$ ways to distribute these letters. But I know that this isn't done yet as I haven't accounted for the relative positions of the letters once placed in the gap(We need to permute that also). 
So I wrote down the various distributions possible of number of letters in each gap.
$(3+1+1+1+1)$ type $\rightarrow 3! \cdot 5!/4!$ permutations
$(4+1+0+1+1)$ type $\rightarrow 4 ! \cdot 5!/3!$ permutations.
$5+1+1+0+0$ type $\rightarrow 5! \cdot5!/(2!\cdot2!)$ permutations.
$(6+1+0+0+0)$ type $\rightarrow 6 ! \cdot 5!/3!$ permutations.
$(7+0+0+0+0)$ type $\rightarrow 7 ! \cdot 5!/4!$ permutations.
$2+2+3+0+0$ type $\rightarrow 2 ! \cdot2!\cdot3!\cdot 5!/(2!\cdot2!)$ permutations.
$2+3+1+1+0$ type $\rightarrow 2 ! \cdot3!\cdot 5!/(2!)$ permutations.
$2+4+1+0+0$ type $\rightarrow 2 ! \cdot4!\cdot5!/(2!)$ permutations.
$2+5+0+0+0$  type $\rightarrow 2 ! \cdot5!\cdot 5!/(3!)$ permutations.
$3+3+1+0+0$ type $\rightarrow  3! \cdot3!\cdot 5!/(2!\cdot2!)$ permutations.
$3+4+0+0+0$ type $\rightarrow 4!\cdot3!\cdot 5!/(3!)$ permutations.
$2+2+2+1+0$ type $\rightarrow 2!\cdot2!\cdot2!\cdot 5!/(3!)$ permutations.
$2+2+3+0+0$ type $\rightarrow 2!\cdot2!\cdot5!/(2!\cdot2!)$ permutations.
Adding all these up and multiplying the sum with $5^7$ I get some huge answer which is wrong :(
Obviously there was to be another way to solve this but I couldn't think of any.

The capital letters can be arranged in $6!=720$ ways.  The lowercase letters can be arranged in $\binom{3}{1}=3$ ways, since we are forced to have x's at either end.
With that done, the capital letters and lowercase letters can be merged in $\binom{11}{5}=462$ ways -- all we need to do is choose which five spots out of the 11 will have lowercase letters since we have already determined the order in which they appear.
$720\cdot3\cdot462=997920$.
