How many four letter words having exactly two vowels can be formed by using the letters of the word "TAILOR"?

Question

(No letter appears twice in the word formed)

Here why can't we use $3\cdot2\cdot3\cdot2\cdot4!$ instead of $\binom{3}{2}\cdot\binom{3}{2}\cdot4!$

Answer 1

Take two vowels out of the three, that is $\binom32=3$ choices. Take other two letters out of the three remaining, that is $\binom32=3$ choices.

Multiply $3\cdot3=9$ to get the different choices of $4$ letters.

Order each choice of $4$ letters in every possible way, that is $9\cdot4!=9\cdot24=216$.

Disclaimer: this way you count "words" regardless of their "meaning" if any. This includes words like AORL, ITRO and such.

Answer 2

When you write $3 \cdot 2$ you're essentially taking a variation, that is, you're taking the vowels but also assigning an order to them, saying "one of these is going to go first, and the other one later". So if you take $3 \cdot 2 \cdot 3 \cdot 2$ ways of choosing two vowels and two consonants, you have already chosen the relative order between consonants and the relative order between vowels

What you ought to do is find the ways of ordering such a word if the order for vowels and consonants is already chosen. This can be done by replacing both vowels with a symbol that represents both, and both consonants with another symbol that does the same. Then, you find the ways of ordering a 4-letter word composed by two symbols, each repeated twice. This number is $\frac{4!}{2! 2!} = 3!$.

You can check that $3 \cdot 2 \cdot 3 \cdot 2 \cdot 3! = {3 \choose 2} \cdot {3 \choose 2} \cdot 4!$