Answer:
There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400