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.
Best Answer
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.