How many ways can SLUMGULLION be arranged so all three L's precede all other consonants? How many ways can the letters in the word SLUMGULLION be arranged so that the three L's precede all the other consonants.
Attempt: There are 11 letters, and there are 3 Ls, 4 vowels: U U I O, and 4 consonants: S M G N.
Then Ls can be arranged in 3!, vowels in 4!/2!, and consonants in 4! ways.  Let V = vowel, L = L, and C = consonant. The number of ways of for L to be before all other consonants are the possible combinantions VLC, LVC, LCV. Thus there are 3. Then we multiply 3(3!*4! *4!/[2!]) is this correct?
Thank you for any feedback.
 A: First we make a $7$-letter sequence using only consonants, with the L's in front. There are $4!$ ways to do this.   Leave fat gaps between successive consonants, we may want to slip vowels  between them.
The I can be placed in our pure consonant sequence in $8$ ways.
Then the O can be inserted in the resulting sequence in $9$ ways.  
The U's are more tricky. Either we use UU, which can be placed in $10$ ways, or we choose $2$ of the $10$ "gaps" in the $9$-letter sequence we have so far, to slip a U into. So the U's can be placed in $10+\binom{10}{2}$ ways.
A: EDIT: I was using the spelling originally given in the title, rather than in the statement of the problem; so I have edited my answer.
There are $\displaystyle\frac{11!}{3!2!}$ total ways to arrange the letters of SLUMGULLION, and 
there are $\displaystyle\frac{7!}{3!}=840$ ways to arrange the consonants L,L,L,S,M,G,N. $\;\;$ There are $4!$ of these arrangements with the L's in front of the other consonants (since we have LLL_ _ _ _ with $4!$ ways to arrange S,M,G,N)
so there are $\displaystyle\frac{4!}{840}\left(\frac{11!}{3!2!}\right)=95,040$ possible ways to do this.

Alternatively, first arrange the 7 consonants L,L,L,S,M,G,N in order with the L's in front; as above, there are $4!$ ways to do this.  
Next we can place 4 spaces for the vowels between these letters, so there are
$\dbinom{11}{4}$ ways to do this ${\hspace.3 in}$(since there are 4 spaces and 7 dividers).
Finally, we can arrange the vowels U,U,I,O in these spaces in $\displaystyle\frac{4!}{2!}=12$ ways;
so we have $4!\dbinom{11}{4}\cdot12=95,040$ possible arrangements.
