Q:

How many 10 letter 'words' can be made with no repeated letters and at least 8 consonants?

Need a little bit of help with this one.
How many 10 letter 'words' can be constructed which have no repeated letters and at least 8 consonants?
where 'word' is defined as any combination of the 26 letters of the english alphabet, and by consonant, (obviously) a letter of the english alphabet that are not either a,i,u,e, or o.
This is where I am at:

We need to choose the positions for
the 8 consonants which is $c(10,8)$
Then we need to choose the 8
consonants which I think is
$p(21,8)$? (not sure about this one)
Choose the remaining 2 letters: We have already used 8 letters so we only have 18 letters to choose from so $p(18, 2)$

What else needs to be calculated? Do we need to choose the position of the remaining 2 letters?

A:

I think the best way would be to separate it into 3 non-overlapped groups:

Words with exactly 8 consonants
Words with exactly 9 consonants
Words with exactly 10 consonants

Logic would be similar with Arturo. Arvin comments wouldn't work, because you count same words several time. Once when the consonant is one of the 8, and again, when it happens to be one of the two remaining letters.
For the 8 consonants:

Select 8 positions - C(10,8)
Select the 8 consonants out of the 21, in order - P(21,8)
Select the 2 vowels in order - P(5,2)

So answer is: 
c(10,8) * p(21,8) * p(5,2) = (10!/(2!*8!)) * (21!/13!) * (5!/3!) = 45 * (21!/13!) * 20
Similarly, do the same for 9 and 10:

9 consonants:  C(10,9) * P(21,9) * P(5,1) = 10 * (21!/12!) * 5 = 50 * (21!/12!)
10 consonants: C(10,10) * P(21,10) * P(5,0) = (21!/11!)

Sum the 3 answers.

A:

Your argument has a couple of problems:

You don't specify where the last two letters selected go; you pick the two letters, but you never place them; placing them in different order gives you different solution.
Likewise, you never say how you will distribute the 8 consonants you pick; you select the 8 locations, but you don't say which letter goes where.
Finally, your solution counts some words too many times. Say you first select the eight consonants b,c,d,f,g,h,j,k and the first eight positions for them (in that order). Then you select two more letters, say m,n, and place them in order in positions nine and ten. You will "count" this word again if your eight selected consonants are b,c,d,f,g,h,m,n, the eight positions selected are positions one through six, nine, and ten (in that order); and then the other two letters you select are j and k and you place them in positions seven and eight. You will count this word several times through your process.

So: perhaps it's better if we start by counting how many 10 letter strings have exactly eight consonants and no repeated letters. My recommendation:

First select the eight positions that will contain the consonants.
Then select the consonants and place them in those positions in order (that is, which consonant you select first matters, because that will go in the leftmost selected position). 
Finally, select the two vowels that will fill the remaining two positions, again in order. 

Then count how many of your strings have exactly 9 consonants in a similar manner; then how many have all letters consonants (that will be the simplest one).

