raju has forgotten his six digit id no. he remembers the following the first two digits are either 1,5 or 2,6 the no. is even and 6 appears twice.if raju uses trial and error process to find his ID no. at the most, how many trials does he need to suceed

Hi Gaurav, 1: 1, 5, _, _, _, _ There are two cases possible. If the unit digit is 6 and if the unit digit is not 6. (i) If the unit digit is 6 1, 5, _, _, _, 6 6 can be placed in any of the three places, and the remaining can be arranged in 9*9 ways. Therefore, the total number of ways = 3*9*9 = 243 (ii) If the unit digit is not 6 1, 5, _, _, _, _ The last digit will have 4 possibilities. (0,2,4,8) For the remaining digits, 2 digits will be 6, and the other can be any number other than 6. Therefore, the total number of ways = 3*9*4 = 108 2. 2, 6, _, _, _, _ (i) If the last digit is 6 2, 6, _, _, _, 6 Total number of ways = 9*9*9 = 729 (ii) If the last digit is not 6 2, 6, _, _, _, _ The last digit has 4 possibilities, and the remaining numbers can be arranged in 3*9*9 ways. Therefore, the total number of ways = 4*3*9*9 = 972 Maximum number of trials required = 108 + 243 + 729 + 972 = 2052 Hope this helps!