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+6−1C6−1=8C5=8C3=56