probability stuff.

[pleasehelp.dontunderstand]

Hello, I am stuck on part of a question and I have tried several methods to solve it and I want to see if anyone can see if I am on the right track or if help me get there...

Okay the question is --> (note: this is not actually the question. I have changed the theme, the wordings and the numbers as I don't want my homework question floating around the internet, but I'm hoping it will still give off the general idea. Also I am allowed to work with people on homework so you aren't helping me cheat!)
You have a jar of eighteen tokens. Three of the tokens are gold, six of the tokens are silver and nine of the tokens are bronze. Using a completely random method, three tokens are chosen and each token is replaced directly after it was chosen. What is the probability of having selected exactly one gold token, one silver token and one bronze token (in any order)?

What I have done:
First I have tried to find the total number of different ways that 3 tokens can be selected with replacement.
(n + k - 1) (18 + 3 - 1)
( k ) ( 3 )
=
(20!)/((20-3)!(3)!) = 1140

And then next I have tried to find the total number of ways to draw each exactly one of each color and I have multiplied them together
For gold with replacement,
(n + k - 1) (3 + 1 - 1)
( k ) ( 1 )
=
(3!)/((3-1)!(1)! = 3

For silver with replacement,
(n + k - 1) (6 + 1 - 1)
( k ) ( 1 )
=
(6!)/((6-1)!(1)! = 6

For bronze with replacement,
(n + k - 1) (9 + 1 - 1)
( k ) ( 1 )
=
(9!)((9-1)!(1)!)) = 9

And now, total ways to get exactly chose exactly one of each...
3 * 6 * 9 = 162

And lastly, to find the probability I divide the ways to get exactly one of each by the total ways...
162/1140 which simplifies to 27/190


^^ I came up with that using a class example problem, but something just feels off about it...
I'd greatly appreciate any help or suggestions!

 

[Harry_the_cat]

I would approach it another way.
Consider the prob of getting GSB in that order (with replacement).
P(G then S then B)= P(G)*P(S)*P(B) because they are independent because of replacement
= \(\displaystyle \frac{3}{18}*\frac{6}{18}*\frac{9}{18}=\frac{1}{36}\)

BUT the order doesn't matter.
There are 3!=6 different orders possible: GSB, GBS, SGB, SBG, BGS, BSG

So P(GSB in any order) is \(\displaystyle 6*\frac{1}{36}= \frac{1}{6}\)

 

[pleasehelp.dontunderstand]

Hmm, is it really that simple? It makes sense to me, but it seems odd since all the other questions and use the permutation/combination formulas. ??‍
Anyways, thanks for your response!!

 

[Harry_the_cat]

There's usually more than one way to do these problems. I often find them easier if you go back to basics and forget the formulas.