**Question:**

Find the number of ways of selecting 9 balls from 6 red balls, 5 while balls and 4 blue balls if each selection consists of 3 balls of each colour.

**Solution:**

Total number of red balls = 6

Total number of white balls $=5$

Total number of blue balls $=4$

No. of ways of selecting 3 balls which is red $={ }^{6} \mathrm{C}_{3}$

No. of ways of selecting 3 balls which is white $={ }^{5} \mathrm{C}_{3}$

No. of ways of selecting 3 balls which is blue $={ }^{4} \mathrm{C}_{3}$

Thus, by Multiplication principle, the total number of ways would be,

$\Rightarrow{ }^{6} \mathrm{C}_{3} \times{ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{3}$

Applying formula, ${ }^{\mathrm{n}} \mathrm{C}_{r}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$, we get

$\Rightarrow 800$ ways

Thus, the number of ways of selecting 9 balls from 6 red balls, 5 while balls and 4 blue balls if each selection consists of 3 balls of each colour would be 800 .