**Toolbox:**

- No. of arrangement of n things out of which p are of one category and q are of another category is $\large\frac{n!}{p!.q!}$

Since no conditions for white and black balls,

2 white and 3 black balls can be arranged in $\large\frac{5!}{2!.3!}$$=10 \:ways$

Since no two red can come together, red balls are to be placed

in between these 5 balls .(2 white,3 black)

There are 6 places and 4 balls as given below

$ - B_1 - B_2 - B_3 - B_4 -B_5 -$

Selection of any 4 places out of 6 places is $^6C_4=15\:ways$

$\therefore$ The required no. of arrangements $= 10\times 15=150$