All defective products must be detected in all six inspections, so the sixth inspection must be one of the three defective products. In addition, two defective products and three genuine products were definitely detected in the first five tests, and one of the three defective products was put into the sixth test. There are three possibilities (C(3, 1) and two possibilities (C(6, 2)). The test results of the first five tests are not limited and can be arranged at will, and there is one possibility (A(5, 5)).

So a * * * has C (3,1) C (6,2 2) A (5 5,5) = 45×120 = 5400 kinds.