Aperson carries 3 bottles of compound x, 3 bottles of compound y and 3 bottles of compound z. while x and y will react vigorously, z will prevent the reaction from happening. if the person falls and 3 random bottles react, what is the probability that a vigorous reaction will happen?

i tried solving this question like this:
possible combinations:
abb
aba

3/6 x 3/5 x 2/4 = 18/120
3/6 x 3/5 x 2/4 = 18/120

18/120 x 2 = 3/10

i'm wondering if the order of the bottles matter and if the the number of bottles of z should be counted (e. g. 3/9 x 3/8 x 2/7)?

tell me if i'm wrong and provide the right way to do

 The number of permutations taken 3 at a time would be

  9! / (3! * 3! * 3!) = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (6 * 6 * 6)

= (9 * 8 * 7  * 5 * 4 * 1)  / 6 = 8 * 7 * 5 * 6 = 1680

Example:  (1 , 2 , 2 ) = 1, 2, 2    2, 1, 2 and 2, 2, 1   3! / 2! = 3

because the 2's are indistinguishable

Now you can put these into groups of 3 and for no reaction all

of the Z bottles must be in 1 group or else a reaction will occur

The number of combinations taken 3 at a time will be

C (9 , 3) = 9! / (3! * 6!) = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (3! * 6)

9 * 8 * 7 / 3! = 12 * 7 = 84    groups taken 3 at a time where all of the Z

bottles are in 1 group.

Then the probability of no reaction is 84 / 1680 = .05

and the probability of a reaction is .95

neither i nor ii 56 / 4.9 = 11.91489362 and 82 / (11+5) =5.125 so i is wrong

ii. (24/8+2) times 3 = 15 and (42+9)/2 =25.5

step-by-step explanation:

56