Solution:

Suppose five friends are abcde and five letters are ABCDE.

According to the conditions, every friend has not received his letter, that is, the same uppercase and lowercase letters cannot be combined.

First, the primary school mathematics calculation method

Analysis process:

Suppose the first letter is a (the order is random and the calculation result has nothing to do with it).

In the first step, A can't blend into A, so A has four possible BCDE—— x 4 (assuming A chooses D).

The second step is to let D choose, and D has four choices of ABCE.

(1) If D chooses A..bce, there are only two arrangements. b-C，c-E，E-B； English, English, Chinese, Chinese-+2

(2) If D chooses one of BCE-X 3 (assuming D chooses E).

Step three, let E choose now. E has three choices, ABC.

If e chooses A..bc, there is only one arrangement. Boston, Boston.—+1

(2) If E chooses one of BC-X 2 (assuming B is chosen).

The fourth step, finally let B choose, B now only has two choices of AC. Actually, because C can't choose C, B can only choose C and let C choose A. So the choice is unique.

To sum up, according to the way of primary school mathematics, the formula is deduced backwards.

[(2x 1+ 1)x3+2 ] x 4 = 44

Second, the enumeration method

a-B，b-A，c-D，d-E，e-C

a-B，b-A，c-E，d-C，e-D

a-B，b-C，c-A，d-E，e-D

a-B，b-C，c-D，d-E，e-A

Arabic, Arabic, Chinese, English, German, English

Arabic, Arabic, Chinese, English, English

A-Ba-Ba-Di, Zhong-Ying, Zhi-Zhong, Ying-A

A -B, b-D, c-A, d-E, e-C

A -B, B-e-C-D, d-C, e-A

A -B, b-E, c-D, d-A, E-C.

a-B，b-E，c-A，d-C，e-D

The above is all the permutations of A and B, totaling 1 1. Select the same number of CDEs.

Therefore, there are 4× 1 1 = 44 species in total.

Third, in addition, chart method, sequential arrangement method and combination method can all be calculated.