Number of permutations under restrictions I need to find how many permutations are there of the letters in:
eeeddnntsc such that beside each n there is d
well i tried with a warm up saying that total permutations for the word:
$$ \left(\frac{10!}{2!*2!*3!}\right) $$
lets assume that i stick dn together, which means we now have 7 letters.
the total permutations: 
$$ \left(\frac{8!*2^2}{2!*3!}\right) $$
the $2^2$ is for ordering the two pairs of dn and the 3! and 2! are for avoiding duplications.
I still however, unhappy with the outcome.
any thoughts?
much appreciation

There are three $e's$, two each of $d$ and $n$, and one each of $s,c,t$.
Leave out the two $d's$ for the moment
There are a total of $\dfrac{8!}{3!2!} = 3360$ permutations with $\dfrac{7!}{3!} = 840$ having the $n's$ together 
In the former case, there are $2$ places for each $d$, thus $(3360-840)\times 4$ ways
In the latter case, one $d$ must be placed in between the two $n's$,
and the other has $10$ places, thus $840\times 10$ ways
Add up.  
