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I assume that with n(...) you mean the size of the set. It sure can; but it need not be.
It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n where nCr = n!/[r!*(n-r)!]
The power of a quotient is the quotient of the power! (a/b)^n = (a^n) / (b^n) where a/b is the quotient and n is the power.
The geometric mean of 'A' and 'B' is the square root of ( A x B ). Edit: The geometric mean of n positive numbers, x1, x2, ... , xn is the nth root of x1*x2* ... *xn.
#include<iostream.h> void main() { cout<<' '<<' '<<"a"<<'\n'; cout<<' '<<"a"<<"b"<<"a"<<'\n'; cout<<'a'<<'b'<<'c'<<'b'<<'a'<<"\n"; }