The number of ways 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is:
Answer: 56
Solutions · 1
Official solution
There are 5 red cubes and 11 blue cubes. Place the blue cubes to create gaps: with 6 gaps around/between the 5 red cubes (positions x1,…,x6), where the 4 internal gaps need ≥2 blue cubes each, and end gaps need ≥0:
\[x_1+x_2+\cdots+x_6=11,\quad x_1,x_6\ge0,\ x_2,x_3,x_4,x_5\ge2\]
Substituting xi′=xi−2 for i=2,3,4,5: sum becomes 11−8=3 distributed among 6 non-negative variables:
\[\binom{3+6-1}{6-1} = \binom{8}{5} = 56\]