算法题:从{1,2,3...N}个元素组成不同的子集合,写不出来感觉有递归和循...答:一个集合含有n个元素,则称它为n元集。一个n元集的子集有2^n。(0)零元集,即空集,有C(n,0)个。(1)一元集:有n个。(2)二元集:有C(n,2)个。...(k)k元集:有C(n,k)个。...(n)n元集:有C(n,n)个。总共有 C(n,0)+C(n,1)+...+C(n,k)+...C(n,n)=2^n ...
若集合A中有n个元素,则集合A的所有不同的子集个数为多少???答:{a1,a2,a3,a4}的子集:φ,{a1},{a2},{a3},{a4},{a1,a2},{a1,a3},{a1,a4},{a2,a3},{a2,a4},{a3,a4},{a1,a2,a3},{a1,a2,a4},{a1,a3,a4},{a2,a3,a4},{a1,a2,a3,a4}【16个=2^4】故猜想集合{a1,a2,a3 ...an}的子集的个数为2^n个 如果不懂,请Hi我,祝学习...