To answer your question, if you use a D2T2 for the "D2", you will have to find ANOTHER cache for the "T2". Otherwise the task could be achieved with a diagonal (9 caches) - and that`ll make it too easy :)
So 18 squares will need to be filled in the matrix. (...Maybe I should have said that straight off? Sorry)
PS: That was actually the problem with the mentioned checker, It counts the same cache for both.