ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

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.

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

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4P3 ways and 4 constants can be arranged in 4P4 ways.

Number of words =4P3  x 4P4= 24 x 24 = 576

There are 7 letters in the word ‘Bengali of these 3 are vowels and 4 consonants.

Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.

Total number of words = 5! x 3!= 120 x 6 = 720

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 word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.