Question

7. 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

39200   are he 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.​

Step-by-step explanation:

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.​

Red Balls can be drawn in ⁶C₃   = 20 ways

Green Balls can be Drawn in ⁸C₃   = 56 ways

Blue Balls can be Drawn in ⁷C₃   = 35 ways

Total number of Ways =  20 * 56 * 35

= 39200

Learn More :

probability that the balls are different

https://brainly.in/question/9675341

Out of 8 different balls taken three at a time

https://brainly.in/question/12318824

Answer 2

Answer:  The required number of ways is 39200.

Step-by-step explanation:  Given that a bag contains 6 red balls, 8 green balls and 7 blue balls.

We are to find the number of ways in which 9 balls can be drawn so that there are 3 balls of every color.

The number of ways in which 3 red balls can be drawn from the 6 red balls in the bag is

$n_1=^6C_3=\dfrac{6!}{3!(6-3)!}=\dfrac{6\times5\times4}{3\times2\times1}=20.$

The number of ways in which 3 green balls can be drawn from the 8 green balls in the bag is

$n_2=^8C_3=\dfrac{8!}{3!(8-3)!}=\dfrac{8\times7\times6}{3\times2\times1}=56.$

The number of ways in which 3 blue balls can be drawn from the 7 blue balls in the bag is

$n_3=^7C_3=\dfrac{7!}{3!(7-3)!}=\dfrac{7\times6\times5}{3\times2\times1}=35.$

Therefore, the number of ways in which 9 balls can be drawn so that there are 3 balls of every color is

$n=n_1\times n_2\times n_3=20\times56\times35=39200.$

Thus, the required number of ways is 39200.