例:S={1·a1a2…,1a},求r排列数 解:设排列数为{pn},gn(x)=1+x,则 g(x)=(1+x)=∑C(m,r)x7 ∑ r=0 !(n-r) n! x r=o(n-r 所以p=n!(a-r)!=p(n,r)例：S={1·a1 ,1·a2 ,…,1·ak },求r-排列数 解：设排列数为{pr }, gri(x)=1+x，则    = = = − = − = = + = k r r k r r k r k r r x n r n x r n r n g x x C n r x 0 0 0 ( )! ! ! !( )! ! ( ) (1 ) ( , ) 所以pr=n!/(n-r)!=p(n,r)