P(4,3)= P34= 24

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.

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

This can be done in 11C4 ways = 330 ways

Three digit number will have unit’s, ten’s and hundred’s place.

Out of 5 given digits any one can take the unit’s place.

This can be done in 5 ways. ...              (i)

After filling the unit’s place, any of the four remaining digits can take the ten’s place.

This can be done in 4 ways. ...              (ii)

After filling in ten’s place, hundred’s place can be filled from any of the three remaining digits.

This can be done in 3 ways. ... (iii) 

So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

Having chosen this, the second letter can be chosen in 26 ways

The first two letters can chosen in 26 x 26 = 676 ways

Having chosen the first two letters, the third letter can be chosen in 26 ways.

All the three letters can be chosen in 676 x 26 =17576 ways.

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

The given statement is clearly true.