QJEEM23Maa607: JEE Main 2023 Mathematics - Class 11 JEEM23Maa607 Solution: If the letters of the word MATHS are permuted and all possible…

To solve this problem, we start by finding the number of permutations before we reach a word that begins with T.

We have 5 letters in the word MATHS. If we fix the first letter, there are 4! (=4$$\times$$3$$\times$$2$$\times$$1=24) ways to arrange the remaining letters.

The letters before T in alphabetical order are A, H, M, and S. For each of these 4 letters, there are 4! ways to arrange the rest of the word, giving 4$$\times$$4! = 96 words that come before any words starting with T.

Then, in the section of words that start with 'T', we start with 'TA'. There are 3 remaining letters after 'TA', hence the permutations for words that start with 'TA' are (5-2)!=3!=6.

So, the total permutations for all words that come before 'TH' is 96 + 6 = 102.

Now, we need to count the words that come before 'THAMS'. After 'TH', the letters left are 'AMS'. Arranged in dictionary order, they would be 'AMS', 'ASM', 'MAS', 'MSA', 'SAM', 'SMA'. So, 'THAMS' is the first word in the 'TH' category, so it is the 103^rd word overall when counting from the beginning.

So, the correct answer is 103, which corresponds to Option A.