Kabuki

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.

Hence, the total number of ways = 4! × 5!

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.

But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.

Since the word hexagon contains 7 different letters,the number of permutations is 7P3 = 7 x 6 x 5 =210

First we need to choose two vowels 3C2 and then three consonants 5C3. Now that we have 5 letters required to make the word,arrange them in 5! ways.

So,3C2*5C3*5! = 3600

Total number of alphabets = 10

so ways to arrange them = 10! 

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps 

Hence number of arrangements = 10!/4! x 3! x 2! = 12600

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.