Question

Find the number of ways of drawing 9 balls from a bag that has 6 red balls,
8 green balls and 7 blue balls so that 3 balls of every colour are drawn.​

Answer 1

Answer:

39200

Step-by-step explanation:

Total number of red balls= 6

Total number of green balls = 8

Total number of blue balls = 7

Total balls to be drawn is 9

Condition is that 3 balls of each colour should be drawn.

Applying Combination :-

Ways of drawing the balls from the set will be as follows :-

${}^{6}c \frac{}{3} \times {}^{8}c\frac{}{3} \times {}^{7}c\frac{}{3}$

Now calculating the above expression:-

$= \frac{6!}{3!\times 3!} \times \frac{8!}{5! \times 3!} \times \frac{7!}{4! \times 3!} \\ \\ \\ = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \\ \\ \\ = 5 \times 4 \times 8 \times 7 \times 7 \times 5 \\ \\ \\ = 20 \times 40 \times 49 \\ \\ \\ = 100 \times 8 \times 49 \\ \\ \\ = 39200$

$\rule{200}{4}$

Answer 2

Answer:

39200

Step-by-step explanation:

Total number of red balls= 6

Total number of green balls = 8

Total number of blue balls = 7

Total balls to be drawn is 9

Condition is that 3 balls of each colour should be drawn.

Applying Combination :-

Ways of drawing the balls from the set will be as follows :-

$\begin{gathered}{}^{6}c \frac{}{3} \times {}^{8}c\frac{}{3} \times {}^{7}c\frac{}{3} \\ \\ \\ \\\end{gathered}$

Now calculating the above expression:-

$\begin{gathered}= \frac{6!}{3!\times 3!} \times \frac{8!}{5! \times 3!} \times \frac{7!}{4! \times 3!} \\ \\ \\ = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \\ \\ \\ = 5 \times 4 \times 8 \times 7 \times 7 \times 5 \\ \\ \\ = 20 \times 40 \times 49 \\ \\ \\ = 100 \times 8 \times 49 \\ \\ \\ = 39200\end{gathered}$

$\rule{200}{4}$