A box contains three red balls,two green balls and one blue ball.In how
Many ways can two ball be chosen?
28
21
To find the number of ways two balls can be chosen from a box containing three red balls, two green balls, and one blue ball, we can use the concept of combinations.
In this problem, we need to choose two balls, regardless of the order. This is represented by the combination formula, which is given by:
nCr = n! / (r! * (n - r)!)
Where n is the total number of balls in the box and r is the number of balls to be chosen.
In this case, n = 6 (3 red + 2 green + 1 blue) and r = 2 (choose two balls).
Applying the formula, we have:
6C2 = 6! / (2! * (6 - 2)!)
Now, let's calculate the values:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
2! = 2 * 1 = 2
(6 - 2)! = 4!
4! = 4 * 3 * 2 * 1 = 24
Substituting these values:
6C2 = 720 / (2 * 24)
Simplifying:
6C2 = 720 / 48
6C2 = 15
Therefore, there are 15 ways to choose two balls from the given box.
Really? Just list them:
RR, RG, RB
GG, GR, GB
BR, BG