There are a total of $6$ red balls, $5$ white balls, and $4$ blue balls.

$9$ balls have to be selected in such a way that each selection consists of $3$ balls of each colour. Here,

$3$ balls can be selected from $6$ red balls in $^{6} C_{3}$ ways.

$3$ balls can be selected from $5$ white balls in $^{5} C_{3}$ ways.

$3$ balls can be selected from $5$ blue balls in $^{5} C_{3}$ ways.

Thus, by multiplication principle, required number of ways of selecting $9$ balls.

$=^{6} C_{3} \times^{5} C_{3} \times^{5} C_{3}=\frac{6 !}{3 ! 3 !} \times \frac{5 !}{3 ! 2 !} \times \frac{5 !}{3 ! 2 !}$

$=\frac{6 \times 5 \times 4 \times 3 !}{3 ! \times 3 \times 2} \times \frac{5 \times 4 \times 3 !}{3 \times 2 \times 1} \times \frac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1}$

$=20 \times 10 \times 10=2000$