High school mathematics c formula
(1) Several knowledge points: (! )C(m,n)=n! /[m! *(n-m)! ].(! ! )(m+ 1)! =(m+ 1)*m! (2) Prove that because C(m, n)=n! /[m! *(n-m)! ]=[(n+ 1-m)*n! ]/[m! *(n+ 1-m)! ].C(m- 1,n)=n! /[(m- 1)! *(n+ 1-m)! ]=[m*n! ]/[m! *(n+ 1-m)! ]. So the right side = c (m, n)+c (m- 1, n) = {[(n+ 1-m) * n! ]/[m! *(n+ 1-m)! ]}+{[m*n! ]/[m! *(n+ 1-m)! ]}={n! *[(n+ 1-m)+m]}/[m! *(n+ 1-m)! ]=(n+ 1)! /[m! *(n+ 1-m)! ]. Obviously, the left =C(m, n+ 1)=(n+ 1)! /[m! *(n+ 1-m)! ]. So left = right, which means c (m, n+ 1) = c (m, n)+c (m- 1, n).