As long as there is one method, this beat combination should be listed one by one, the first one starts with a, and there is ABCDE. ABCE。 ABEC and other AB. Advertising. AE3× 3 = 9。

Now, for example, badec.badce at the beginning of b has ba.bc.bd.be2× 4 = 8 respectively.

So there are eight kinds of 3× 8 = 24 at the beginning of b.d.e, and 2× 9 = 18 at the beginning of A.C. So the answer is 24+ 18 = 42.