How many arrangements of the letters in the word CALIFORNIA have no consecutive letter the same?
Solution 1:
We modify your approach by first placing the A's, then placing the I's.
We can arrange the six distinct letters C,L,F,O,R,N in $6!$ ways. This creates seven spaces, five between successive letters and two at the ends of the row.
Case 1: We choose two of these seven spaces in which to place the two A's, thereby separating them.
We now have eight letters. This creates nine spaces, seven between successive letters and two at the ends of the row. We choose two of these nine spaces for the I's. The number of such arrangements is $$6!\binom{7}{2}\binom{9}{2} = 544,320$$
Case 2: We place both A's in the same space.
We again have eight letters. This again creates nine spaces. The space between the two A's must be filled with an I. Therefore, there are eight ways to choose the position of the other I. The number of such arrangements is $$6!\binom{7}{1}\binom{8}{1} = 40,320$$
Total: These two cases are mutually exclusive. Hence, the total number of arrangements of the letters of the word CALIFORNIA in which no two consecutive letters are the same is $$6!\left[\binom{7}{2}\binom{9}{2} + \binom{7}{1}\binom{8}{1}\right] = 584,640$$ which agrees with the result obtained by using the Inclusion-Exclusion Principle.
Solution 2:
$\underline{A\; shorter\; method}$
Suppose $2$ objects are to be kept apart in a permutation, there are $2$ well known methods
the "gap method", where these $2$ objects are placed in the gaps (incl. ends)
the "subtraction method" [ total permutations - those with the $2$ objects together ]
Find the number of ways both $A's$ and $I's$ are apart by successively applying them
$A's$ apart or together, $I's$ apart $= (8!/2!)\binom92= 725,760$
$ A's$ together, $I's$ apart $= 7!\binom82 = 141,120$
thus both $A's$ and $I's$ apart = $725,760 - 141,120 = 584,640$
Solution 3:
There are just $3$ patterns for first placing the $A's \; and\; I's\;$ some of which need separators glued to first occurrence of the"double(s)" and shown by large letter(s).
Both "doubles" together:
${\Large A}A{\Large I}I\quad or \quad{\Large I}I{\Large A}A$ which use up two of the other letters in $6\cdot5$ ways,
and the remaining $4$ can be successively inserted in $5\cdot 6\cdot 7\cdot 8$ ways
One double together:
$A{\Large I}IA \quad or \quad I{\Large A}AI$, which uses up one letter in $6$ ways
and the remaining $5$ can be successively inserted in $5\cdot6\cdot7\cdot8\cdot9$ ways
No doubles together:
$AIAI\quad or \quad IAIA$
and the remaining $6$ can be inserted in $5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10$ ways
Adding up, we get $2\cdot 5\cdot 6\cdot 7 \cdot 8(6\cdot 5 +6\cdot 9+ 9\cdot 10) = 584,640$
ADDED
I subsequently found a shorter method, which I am posting separately, but leaving this one, too.
Solution 4:
The problem here is that you're not counting the possibility that an $A$ and an $I$ could be in the same "space," like in this string: $$CALIFORNIA$$ Here, we don't have consecutive same letters, but "I" and "A" are both at the end, in the seventh space, which you're not accounting for.
Therefore, we have to consider the $A$s and $I$s separately. We have $6!$ distinct arrangements of the distinct letters and can then put the $A$s in $7 \choose 2$ spots and can then put the $I$s in $7 \choose 2$ spots, so we have: $$6\cdot {7 \choose 2}\cdot {7 \choose 2}=317520$$ This is wrong, however, because we're not accounting for the order of $A$ and $I$ if they are in the same space. If you can figure out how to account for that, you can salvage your method, but I think the inclusion-exclusion is quite easier at this point.