Question 1098609
<pre>
First we find the number of sequences of 4 letters with no
two adjacent letters the same.  There are 25 letters that 
can be used, all the letters except "I".

Case 1.  All 4 letters used are different.
None of these have 2 adjacent letters (or any letters) the same.
That's 25P4 = 303600.

Case 2. Exactly 3 different letters are used.
These have two letters the same. The only positions the two
like letters can occur and not be adjacent are
1st and 3rd, 
1st and 4th, and 
2nd and 4th.  
Examples of sequences of these three forms are ABAC, ABCA, 
and ABCB.  So every sequence must be of one the same 3 forms 
as those 3.
There are 25 ways to put a letter where the A's are.
There remain 24 ways to put a letter where the B's are.
There remain 23 ways to put a letter where the C's are.
That's (3)(25P3) = (3)(25)(24)(23) = 41400

Case 3. Exactly 2 different letters are used.
So they can only be 2 pairs of like letters.
To have no 2 like letters adjacent, one pair must
come 1st and 3rd and the other pair 2nd and 4th.
An example is ABAB.
There are 25 ways to put a letter where the A's are.
There remain 24 ways to put a different letter where the 
B's are.
That's (25)(24) = 25P2 = 600

Total for all 3 cases = 303600+41400+600 = 345600

For each of those 345600 ways to choose the 4 letters,
the 3 digits can be any integer from 100 to 999, inclusive.
There are 999 integers from 1 to 999 inclusive, and we subtract
99 for the 99 integers from 1 to 99 inclusive, and get 
999-99 = 900.

So for each of the 345600 ways to choose the letters, there
are 900 ways to choose the 3 digits.

Answer (345600)(900) = 311040000

Edwin</pre>