1. Interpret the problem

We have $16$ cubes in a row:

• $11$ blue cubes (identical)
• $5$ red cubes (identical)

We need the number of arrangements such that between any two red cubes there are at least 2 blue cubes.

Since cubes of the same color are identical, this is equivalent to counting color patterns made of $11$ B's and $5$ R's.

2. Set up the gap method

Place the $5$ red cubes first:

$R\ \_\ R\ \_\ R\ \_\ R\ \_\ R$

There are $4$ gaps between consecutive red cubes, and each such gap must contain at least 2 blue cubes.

So the minimum number of blue cubes required between the reds is:

$4 \times 2 = 8$

Out of the $11$ blue cubes, after placing these $8$, we have

$11 - 8 = 3$

blue cubes left.

So the structure becomes:

$x_0\, R\, (2+x_1)B\, R\, (2+x_2)B\, R\, (2+x_3)B\, R\, (2+x_4)B\, R\, x_5$

More simply, distribute the remaining $3$ blue cubes into the $6$ possible places:

• before the first red,
• the 4 already-satisfied internal gaps,
• after the last red.

Let these extra blue cubes be $y_1,y_2,y_3,y_4,y_5,y_6 \ge 0$. Then

$y_1+y_2+y_3+y_4+y_5+y_6 = 3$

3. Count non-negative integer solutions

The number of non-negative integer solutions to

$y_1+y_2+y_3+y_4+y_5+y_6=3$

is given by stars and bars:

$\binom{3+6-1}{6-1} = \binom{8}{5} = \binom{8}{3}$

$\binom{8}{3} = \frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1} = 56$

4. Final answer

Hence, the number of valid arrangements is

$\boxed{56}$

5. Comparison with stored answer

Stored correct answer = $56$.

Our derived answer is also $56$, so they agree.