How many 3 letters code can be formed by using the five vowels without repetitions?

Hint: Make blank spaces of 3 sections and mention position in it to ease the statement because repetition is not allowed here and try to make arrangements for the specified positions.

Complete step by step answer:
Given: A code of $3$ letters is to be formed using $5$ vowels.
$5$ vowels in English alphabet $ = $ a, e, i, o, u, as it is given that none of the vowels will be repeated it means if one vowel is placed at one position then only four vowels will be remained for the next two position.
Let us understand by making a small $3$ section box.
                                   ___1st____, __2nd___, ___3rd___
If out of $5$ vowels (a, e, i, o, u), one of the vowels placed at 1st position then only $4$ vowels will be available for the next two positions.
So the vowels available for 1st position $ = 5$ vowels and if placed another vowel out of remaining $4$ vowels at 2nd position 2nd position 2nd position then only $3$ vowels will be available for next 3rd position.
So the vowels available for 2nd position $ = 4$ vowels and for 3rd position vowels available $ = 3$ vowels.
So the total no. of code of $3$ letter can be formed by using five vowels $ = $no. of vowels available for 1st position $ \times $ no. of vowels available for 2nd position $ \times $ no. of vowels available for 3rd position.
The total no. of code of $3$ letters can be formed by using $5$ vowels $ = 5 \times 4 \times 3$
=60 codes

Therefore, Total 60 three lettered codes can be formed by using the five vowels without repetitions.

Note:
As repetition of the vowels is not allowed therefore after every position one vowel will not be available for the next position.
Alternatively, we can use direct formula of choosing items 3 out of 5 as \[{}^5{P_3}.\] And, \[{}^5{P_3} = \dfrac{{5!}}{{(5 - 3)!}} = \dfrac{{5!}}{{2!}}.\]