Probability that the first 2 letters are consonants when the letters of the word 'equilibrium' are rearranged Here's what I tried:
The total number of ways is $\dfrac{11!}{3!\cdot2!}$.
The consonants can be together in $\dfrac{6(5)(9!)}{3!\cdot 2!}$ ways. When I divide, I get $\dfrac 3{11}$ but the answer is $\dfrac2{11}$.... Where did I go wrong?
The second part states find the probability that all the vowels are together. 
I did $7!\times 10$ ($7$ if you consider all vowels as one unit, then multiply by $10$ cause you can rearrange vowels amongst themselves in $10$ ways), divided by the total number of ways, and I got $\dfrac1{66}$... I don't see where I went wrong. [The answer = $\dfrac2{77}$] 

It's all about trying possible paths/combinations and their probabilities. For the first example, proceed sequentially:


*

*Probability that the first letter that you pick is a consonant: $\frac{5}{11}$ (5 consonants over eleven letters)

*Probability that the second letter that you pick is a consonant: $\frac{4}{10}$ (recall that you already picked one). 


The probability you are looking for is $$\frac{5}{11}*\frac{4}{10}=\frac{2}{11}$$
As for the second case, we can reason the same way. However, we will try a different argument. First, note that there are 11! possible combinations of the letters. Likewise, there are 6! ways to combine 6 elements and 5! possible ways to combine 5. Thus, the probability of getting any 6 particular letters together in a combination (for example, the six vowels) is $$6*\,\bigg(\frac{6!\,5!}{11!}\bigg)=\frac{1}{77}$$
The 6 at the beginning of the expression comes from the number of places (in an eleven-slot element) where you can have a sequence of 6 consecutive elements. 
