The number of ways, 16 identical cubes, of which 11 are blue and...

Given, $4$ blue & $5$ red spherical ball

Now according to question there should be minimum 2 blue ball in between any two red ball.

So assume $a, b, c, d, e, f$ are number of blue balls,

So, $a + b + c + d + e + f = 11$ ......(i)

Where $b, c, d, e ⩾ 2$

Now let $b = {b}^{'} + 2$

$c = {c}^{'} + 2$

$d = {d}^{'} + 2$

$e = {e}^{'} + 2$

So equation (i) becomes

$a + {b}^{'} + 2 + {c}^{'} + 2 + {d}^{'} + 2 + {e}^{'} + 2 + f = 11$

$⇒ a + {b}^{'} + {c}^{'} + {d}^{'} + {e}^{'} + f = 3$

Now total ways will be

${=}^{3 + 6 - 1}{C}_{6 - 1}{=}^{8}{C}_{5}{=}^{8}{C}_{3} = 56$