Permutations and Probability question...

I don't think I ever learned about this stuff. I'm trying to take some exams for tutoriing job and there is a topic that's in my way though. I get stuck on problems like this:

*Say I have 6 blue coins, 8 green coins, and 4 red coins, what is the probability that I will draw at random a blue, then a green, then another blue in that order? ....*

So basically from what I've read you can have combinations or permutations, permutations are "orderly" combinations.

Arsenal:

Permutation with repetition where (order matters): $\displaystyle _nP_r=n^r$

Permutation without repetition (order matters): $\displaystyle _nP_r= \frac{n!}{(n-r)!}$

Combination with repetition (order doesn't matter): $\displaystyle _nC_r=\frac{n!}{(n-r)!r!}$

Combination without repetition (order doesn’t matter):$\displaystyle _nC_r=\frac{(n+r-1)!}{r!(n-1)!}$

Where $\displaystyle n$ is the number of things, and $\displaystyle r$ is the number of things being used.

For my case have permuation with repetition and order matters, so is the solution just $\displaystyle \frac{1}{n^r}=\frac{1}{(8+4+6)^3}=\frac{1}{5832}$ ?? that seems a bit off...lol.

__Edit__ Maybe I'm confusing the meaning of "repetition" here. Maybe that does not mean repetition as in the same "type" of coin appearing more than once in a certain permutation, but the repetition of a permutation itself i.e. drawing blue-red-blue could happen more than once.

Re: Permutations and Probability question...

Quote:

Originally Posted by

**adkinsjr** *Say I have 6 blue coins, 8 green coins, and 4 red coins, what is the probability that I will draw at random a blue, then a green, then another blue in that order? ....*

So basically from what I've read you can have combinations or permutations, permutations are "orderly" combinations.

Permutation with repetition where (order matters): $\displaystyle _nP_r=n^r$

Permutation without repetition (order matters): $\displaystyle _nP_r= \frac{n!}{(n-r)!}$

Combination with repetition (order doesn't matter): $\displaystyle _nC_r=\frac{n!}{(n-r)!r!}$

Combination without repetition (order doesn’t matter):$\displaystyle _nC_r=\frac{(n+r-1)!}{r!(n-1)!}$

Where $\displaystyle n$ is the number of things, and $\displaystyle r$ is the number of things being used.

Maybe I'm confusing the meaning of "repetition" here. Maybe that does not mean repetition as in the same "type" of coin appearing more than once in a certain permutation, but the repetition of a permutation itself i.e. drawing blue-red-blue could happen more than once.

The important idea here is with or without replacement. Do we keep the coin or put it back.

If it is without replacement and order makes a difference then $\displaystyle \mathcal{P}(BGB)=\frac{6}{18}\cdot\frac{8}{17} \cdot\frac{5}{16}$.

Now you need to understand, this is not a tutorial service. We cannot offer instruction here. But if you post question along with your effort to solve them, then we can advise you.