A box contains 2 red, 3 black and 4 blue balls. 3 balls are randomly drawn from the box. What is the probability that the balls are of different colors?  Select the correct option
A) 23
B) 25
C) 24
D) 26
E) 29
A box contains 2 red, 3 black and 4 blue balls. 3 balls are randomly drawn from the box. What is the probability that the balls are of different colors?First consider the probability of drawing a red ball at the first draw, a black at the second and a blue at the third. The first probability is 2/9, the second, conditional on the first is 3/8 and the third conditional on the other two is 4/7. The product is (2*3*4)/(9*8*7). It is easy to see that any other order is equally likely (the only difference is the order of the factors in the numerator). So adding 6 equally likely cases we obtain (6*2*3*4)/(9*8*7) = 2/7.   Combinations approach-   Let us assume that all balls are unique.   There are a total of 9 balls.   Total ways = 3 balls can be chosen in 9243 ways = 9(6!3!) = 9*8*7/3*2*1 = 84   Favorable ways = 1 Red ball, 1 Black ball, and 1 Blue Ball = 2*3*4= 
<gadget id="calculator">2*3*4</gadget>
<output>24</output>
24   Answer 24
<result>C</result>
A box contains 2 red, 3 black and 4 blue balls. 3 balls are randomly drawn from the box. What is the probability that the balls are of different colors?