INTERMEDIATE : Not talking about Level. Part-VI

The order of vowels doesn't change => there's only one order => it's as if they were all replaced by one letter, V. So we have 12 letters, out of which 6 are V and 2 are T. $\dfrac{12!}{6! \cdot 2!} = 332640$

Vowels should be in relative order IEEIAE in the word formed . (We can insert any number of consonants between them). There are 12 places to be filled by letters. Number of ways to choose 6 places = 12!/6!6! These 6 places can be filled in only one way that is the order given. The remaining 6 places can be filled in by consonants in 6!/2! ways (as there are two Ts) So total number of ways = 12!6!/6!6!2! That is equal to 332640.😁

Solve using probability. Consider vowels, IEEIAE,number of arrangements is 60(6!÷2!÷3!). probability of this arrangement is 1÷60. So answer is 1÷60×total permutations. That is 1÷60*19958400

Applying the Vandermonde Recursion identity. (n - i)C(r - i -1 ).where i is the place values of the vowels.

I first did the problem thinking it meant that not all letters had to be included but all the vowels did in their relative places and came up with a sum which multiplies the number of ways that the 6 vowels could be placed amongst the $6+n$ letters by the number of unique permutations of $n$ consonants, accounting for the presence of 2 T's: $\sum\limits_{n=0}^6 \frac {(6+n)!}{n! 6!} \left( \frac {5!}{(5-n)!} + \binom {n}{2} \frac {4!}{(6-n)!} \right) = 545952$

If the problem is instead to use all of the letters, we instead simply have to choose the spots for the 6 vowels within the 12 letters and then multiply by the number of unique permutations of the consonants, instead giving an answer of: $\binom {12}{6} \cdot \frac {6!}{2!} = 332640$