Question

Find the number of words those can be formed by using all the letters of the word ‘DAUGHTER’. If:
(i) vowels occurs in first and last place.
(ii) start with letter G and end with letter H.

Answer 1

First, before proceeding for this, we must separate the vowels and consonants from the given word which is ‘DAUGHTER’ as Vowels are ‘AUE’ and consonants are ‘DGHTR’. Then, we can use the concept of factorial for other 6 places which are consonants and the first place has possibility of 3 and last place has possibility of 2, we get the number of ways for the condition. Then, solving for the second condition that is start with letter G and end with letter H, we can see that the two positions are fixed with H and G and no we are left with 6 more letters to be arranged in between left 6 places, we get the number of ways required.

Complete step by step answer:
In this question, we are supposed to find the number of words those can be formed by using all the letters of the word ‘DAUGHTER’ for two conditions as:
(i) vowels occur in first and last place.
(ii) start with letter G and end with letter H.
So, before proceeding for this, we must separate the vowels and consonants from the given word which is ‘DAUGHTER’ as:
Vowels are ‘AUE’ and consonants are ‘DGHTR’.
Then, for solving for the first condition that vowels occur in first and last place.
So, we can see that for the first letter to be vowel, there are three possibilities and if we fill the first letter, then for the last letter we are left with the two possibilities.
Now, we can use the concept of factorial for other 6 places which are consonants and the first place has possibility of 3 and last place has possibility of 2.
The, by applying the above condition we get the number of ways in which vowels occurs in first and last place as:
$3\times 2\times 6!$
Then, by solving the above expression, we get:
$3\times 2\times 6\times 5\times 4\times 3\times 2\times 1=4320$
So, the number of ways in which vowels occur in first and last place are 4320.
Then, solving for the second condition that is start with letter G and end with letter H, we can see that the two positions are fixed with H and G and now we are left with 6 more letters to be arranged in between the left 6 places.
So, to get the number of ways in which start with letter G and end with letter H, we have:
$6!$
Then, by solving the above expression, we get:
$6\times 5\times 4\times 3\times 2\times 1=720$

So, the number of ways in which start with letter G and end with letter H are 720.

Note: Now, to solve these types of questions we need to know some of the basic formulas of factorials beforehand so that we can easily proceed in these types of questions. To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{aligned} & 4!=4\times 3\times 2\times 1 \\\ & \Rightarrow 24 \\\ \end{aligned}$