C(n，m)=n！ /(m！ (n-m)！ )。 It can be seen from this equation that C(n, m)=C(n, n-m). The combination is symmetrical. Interpreted as the total number of different methods to extract m objects from a pile of n objects. On the other hand, if you take out m objects, the remaining n-m objects are also certain, so map f:{n objects take out m objects}->; {n objects take out n-m objects} This is a one-to-one correspondence. So C(n, m)=C(n, n-m).

The second equation is (n-m) c (n, m)+c (n, m- 1) = c (n+ 1), m)? It seems wrong. For example, n=4, m=2, left =2*6+4= 16, right =5*4/2= 10.

C(n+ 1，m)=((n+ 1)n...(n+ 1-m+ 1))/m！ = (n+ 1)/(n-m+1) c (n, m)=C(n, m)+m/(n-m+ 1)C(n, m)=C(n, m).