First of all, you don't distinguish between two boxes, at least put 1 letters.

There may be 1 box A and three boxes B.

Or two boxes of a and two boxes of B.

In fact, it is 4 ÷ 2 = 2.

If it is 5, 5 ÷ 2 = 2.5, round it to 2.

Namely A 1 B4 and A2 B3 (A and B are indistinguishable).

Then do the calculation.

In the case of four letters, there are 1 in box A and three in box B, that is, "C4 1" (I don't know if you have learned A and C, C4 1 refers to randomly taking 1 from four elements, and the calculation formula is cab = (a× (a-6544)

Then there are two boxes A and two boxes B, which means C42 = 6. Because AB is indistinguishable, dividing by 2 is 3. Similarly, when the number of letters in the AB box is the same, this calculation is repeated, so it is divided by 2.

4+3=7

So four letters is seven.

There are 20 letters, that is C201+C20 2+...+(C2010 ÷ 2) = 20+190+... I won't count the rest. .