In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

10’s place can be filled in 10 different ways

100’s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. 

Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

This n3 different initials = 1 million 

i.e. n3=106  (1 million = 106)

  => n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

Required number of ways = (7C5*3C2) = (7C2*3C1) = 63

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

From 5 consonants, 3 consonants can be selected in 5C3 ways.

From 4 vowels, 2 vowels can be selected in 4C2 ways.

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

Total number of words = 5C3*4C2*5P5

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

4 novels can be selected out of 5 in 5C4 ways.

2 biographies can be selected out of 4 in 4C2 ways.

Number of ways of arranging novels and biographies = 5C4*4C2  = 30

After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.

By the Counting Principle, the total number of arrangements = 30 x 720 = 21600