Sequences and Combinatorics, How many arrangements does the word 'FULFILLED' have with following properties I have been given the following word: $FULFILLED$ and have been asked to find the number of arrangements with the following properties:
$\bullet$ No consecutive $F's$
$\bullet$ The vowels $E,I,U$ are in alphabetical order
$\bullet$ The three $L's$ are next to each other
This is what I have so far:
Arranging the vowels in alphabetical order: $E,I,U$, taking up $3$ of the $9$ places
Now we have $2$ $F's$ and $1$ $D$ and $3$ $L's$ (that need to be treated as a group)
There are $6$ places to place the $LLL's$
There are $5$ places left to place the $D$
There are $7\times6$ places left to place the $2$ $F's$, but since they are not distinguishable, this number is divided by $2!$, so it's actually $7\times 3=21$
Thus the number of arrangements with the given properties is:
$7\times3\times6\times5=630$
I know that I went wrong somewhere, but I can't find where.
Can anyone please help me correct my mistake?
 A: Since the three Ls must appear in consecutive positions, we effectively have seven objects to arrange: LLL, F, F, D, E, I, U.
Method 1:  Let's set aside the two Fs for the moment.  That leaves us with five objects to arrange: LLL, D, E, I, U.  There are five ways to place the LLL and four ways to place the D.  Once we do so, there is only one way to arrange the three vowels E, I, and U in the remaining three positions so that the vowels appear in alphabetical order.
Once we have arranged these five letters, we have six spaces in which to place the two Fs, four between successive objects and two at the ends of the row.
$$\square O_1 \square O_2 \square O_3 \square O_4 \square O_5 \square$$
To ensure that the two Fs are not adjacent, we must choose two of these six spaces in which to place an F, which can be done in $\binom{6}{2}$ ways.
Hence, the number of arrangements of the letters of the word FULFILLED in which there are no consecutive Fs, the vowels E, I, U appear in alphabetical order, and the three Ls are consecutive is
$$5 \cdot 4 \cdot \binom{6}{2}$$
Method 2:  Let's initially ignore the condition that the Fs must not be adjacent.
We arrange the seven objects LLL, F, F, D, E, I, U so that the vowels are in alphabetical order.  There are $\binom{7}{2}$ ways to place the two Fs (ignoring the restriction that the Fs are not adjacent), five ways to place the LLL, four ways to place the D, and one way to fill the remaining three positions with the vowels E, I, U so that the vowels appear in alphabetical order.  Hence, if we ignore the restriction that the Fs cannot be adjacent, we obtain
$$\binom{7}{2} \cdot 5 \cdot 4$$
arrangements in which the three Ls are consecutive and the three vowels appear in alphabetical order.
From these, we subtract those arrangements in which the two Fs are adjacent.  We now effectively have six objects to arrange: LLL, FF, D, E, I, U.  There are six ways to place the LLL, five ways to place the FF, four ways to place the D, and one way to arrange the three vowels E, I, U in the remaining three positions so that they appear in alphabetical order.  Hence, there are
$$6 \cdot 5 \cdot 4$$
arrangements in which the three Ls are consecutive, the two Fs are adjacent, and the three vowels appear in alphabetical order.
Hence, the number of admissible arrangements of the letters of the word FULFILLED is
$$\binom{7}{2} \cdot 5 \cdot 4 - 6 \cdot 5 \cdot 4$$
