Let G be a finite group, and ?(G) be the number of conjugacy 2^(| ? (G)|-2). We show that if G is solvable, then ?(G) =2^(| ? (G)|-1) if and only if one of the following holds 1. G ?H ×Z_m,where ?(H) = |?(H)| = 2,is square free and gcd (|H|, m) = 1. 2. G ?H × Z_(r^3 ) × Z_m, where ?(H) = 1, |?(H)| = 2, r is a prime number, m is square free and gcd( |H|,r2m) = 1. 3. G ??a, b, c | a^p= b^(q^2 ) = c^r = 1, [b,c] = 1,b^(-1)ab = a^s, c^(-1)ac = a^t?×Z_m,where p,q and r are different prime number, m is square free and gcd (p q r, m) = 1. 4. G ??a, b, c | a^p = b^q = c^r = 1,[b,c] = 1, b^(-1)ab = a^s,c^(-1)ac =a^t?×Z_m,where p,q and r are different prime number, m is square free and gcd(p q r, m) = 1. 5. G ?Z_p ×Z_q ×Z_mwhere m is square free and gcd (p, m) = 1. 6. G ?Q8 ×Z_m,where m is square free and gcd(2,m) = 1. 8. G ? (Z_p ×Z_p)? Z_m,where m is square free and gcd(p, m) = 1. Also if G is non-solvable, then ?(G) =2^(| ? (G)|-2) if and only if G = H ×Z_m, where H ? ?a, b | a^p = b^(q^m ) = 1, b^(-1)ab = a^s, s ?? 1 (mod p), s^q? 1 (mod p)?, m is square free and gcd(|H|, m) = 1.