Conditional Probability Example using permutations 
You are dealt three cards.  The events of interest concern the number of face cards that you are dealt (0, 1, 2, or 3).  What is the conditional probability that you are dealt at least 2 face cards given that the last card dealt to you was a face card?

There 12 face face cards (1 jack, queen, king * 4 suits) and 40 non face cards. 
Let F= face card, and R = regular or nonface card.
So the question is asking us to find the probability of getting {RFF, FRF, FFF}.  I thought I would approach this question using permutations:
$\Pr(\text{@ least 2 face cards | last card has face}) = \Pr(\{RFF, FRF, FFF\})=\cfrac{(40)(12)(11) + (12)(40)(11) + (12)(11)(10)}{(52)(51)(50)} = 0.089593$
but the answer is 0.3882.  Can anyone please tell me why this is not working?  My understanding is that when you're solving probability you can consider ordering or not consider it.  In other words, you could use permutations or combinations.  
Thank you in advance.
Here is a tree diagram.  Let A, B, and C be events where a face card is obtained.

 A: Last or first or second makes no difference. It will feel clearer with first.
We want the probability of at least one face card out of two, from a deck in which one face card is gone. 
It is easier to find the probability of no face card in two picks. This is $\frac{40}{51}\cdot\frac{39}{50}$.  
So our answer is $1-\frac{40}{51}\cdot\frac{39}{50}$.
Another way: There are fancier ways to proceed. Let $F$ be the event "the third card is a face card," and let $A$ be the event "at least two face cards." We want the conditional probability $\Pr(A|F)$. By the definition of conditional probability, we have
$$\Pr(A|F)=\frac{\Pr(A\cap F)}{\Pr(F)}.\tag{$1$}$$
Note that $\Pr(F)=\frac{12}{52}$. 
You calculated the top of $(1)$ correctly, and did not divide by the bottom. 
Remark: Remember that we are given that the third card is a face card. So the sample space is restricted to those draws in which the third was a face card. What you calculated was the probability that there are at least two face cards and the third is a face card. That is not what the problem asked for, buut once we have found it, the answer is one step away.  
