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!

Hi Gaurav,

Let's consider the 2 cases

1: 1 5 _ _ _ _ 

Now, unit digit can be 0/2/4/6/8.

If unit digit is not 6, it can be selected in 4 ways. But, number must have 2 6s. The third number required can be chosen in 10 ways. So, the total ways will be 3!/2! (arranging 3 digits) X 10 X 4 = 120

If unit digit is 6, other 3 numbers must have exactly 1 '6'. The other 2 numbers can be selected in 10 X 10 = 100 ways. So, number of arrangements will be 3! X 100 = 600

So, this case must have 720 different arrangements.

You can repeat the same process for case 2 as well which is 2 6 _ _ _ _