The number of 6-digit numbers that can be made with the digits 0,1,2,3,4 and 5 so that even digits occupy odd places, is

×|×|×|  The odd place shown by crosses can be filled in 3P3−2P2  ways. (∴ 0 cannot go in the first place form the left). The remaining place can be filled in 3! Ways. ∴ The required number of numbers (3P3−2P2)×3!