Number of ways to place 52 cards in a row the certain way

Suppose we have 52 cards: 26 black and 26 red ones. When the cards are placed in a row if between two given cards of one colour there is no cards of other colour we can take those two out. The questions is in how many ways can we place cards such that we'd be able to take out all cards.

Hint

In general, given $2n$ cards in a row where each is black or red, number the positions they occupy from $1$ to $2n$ from left to right, and consider the quantity $$ (\text{# black cards at an even position})-(\text{# black cards at an odd position}). $$ Whenever you remove two adjacent black cards, you can push the the cards together to get a smaller line of $2n-2$ cards. When you compute the same quantity for the new line, you will find it is the same as before. Same goes for removing two adjacent red cards. So clearly, in order to remove all the cards, it is necessary for this quantity to be zero. It turns out this condition is also sufficient; which I leave to you to prove.

Further hint:

In order to prove sufficiency by induction, it suffices to prove that when there are $2n$ black cards and $2m$ red cards in a line, such that the quantity is zero, then there exists a legal move. This is trivially true when $m\neq n$. When $n=m$, it is easier to prove the contrapositive: "If no legal move exists (and you haven't already won), then the quantity is nonzero."

Once you have proven that a position is solvable if the quantity is zero, you just need to count the number of card placements where there are $13$ black cards at an even position, and $13$ black cards at an odd position.