How many words with letters from the word ABRACADABRA if they must end in a consonant and $d$ must be after $r$. How many words with letters from the word ABRACADABRA if they must end in a consonant and $d$ must be after $r$.
What I did:
I have
$A:5$
$B:2$
$R:2$
$C:1$
$D:1$
If the words must end in a consonant and d must be after r I have only two cases:
1)$D$ at the end.
2)$C$ at the end.
3)$B$ at the end.
Case 1:
$$
\_\  \_\ \_\ \_\ \_\ \_\ \_\ \_\ \_\ \_\ \text{D}  
$$
So I have to choose $10$ letters for the remaining slots with $A$ repeated $5$ times, $2$ $B$s and 2 $R$s:
$$
\frac{10!}{5!2!2!}
$$
Case 2: I set $D=R$ and same thought process as before, giving me:
$$
\frac{10!}{5!3!2!}
$$
Case 3:  Same as case 2.
$$Total= \frac{10!}{5!2!2!}+\frac{10!}{5!3!} $$
Is this correct?

Since D must appear after both R's, the last letter can be a B, C, or D.  
Case 1:  The last letter is D.
We have ten places to fill with five A's, two B's, two R's, and one C.  We can fill five of the ten places with A's in $\binom{10}{5}$ ways.  We can fill two of the remaining five places with B's in $\binom{5}{2}$ ways.  We can fill two of the remaining three places with R's in $\binom{3}{2}$ ways.  Finally, we can fill the last place with a C in $\binom{1}{1}$ way, so there are 
$$\binom{10}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{10!}{5!5!} \cdot \frac{5!}{3!2!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!0!} = \frac{10!}{5!2!2!1!}$$
permutations that end in a D, as you found.  
Case 2: The last letter is C.  
Then we have ten places to fill with five A's, two B's, two R's, and one D.  If, at first, we ignore the requirement that D must appear after the two R's, using the same procedure as above yields
$$\binom{10}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1}$$ 
However, in only one third of these permutations does D appear after both R's.  Thus, the number of permutations in which the last letter is a C and D appears after both R's is 
$$\frac{1}{3}\binom{10}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{1}{3} \cdot \frac{10!}{5!2!2!1!}$$
Case 3:  The last letter is a B.  
Then we have ten places to fill with five A's, two R's, one B, one C, and one D.  If, at first, we ignore the requirement that D must appear after both R's, then using the same procedure as above yields 
$$\binom{10}{5}\binom{5}{2}\binom{3}{1}\binom{2}{1}\binom{1}{1}$$
However, in only one third of these permutations does the letter D appear after both R's.  Thus, the number of permutations in which the last letter is a B and D appears after both R's is 
$$\frac{1}{3}\binom{10}{5}\binom{5}{2}\binom{3}{1}\binom{2}{1}\binom{1}{1} = \frac{1}{3} \cdot \frac{10!}{5!2!1!1!1!}$$ 
To find the number of words that can be formed from the word ABRACADABRA in which the last letter is a consonant and D appears after both R's, add the totals for the three disjoint cases.  
