Question

Find the number of words which begins with a vowel and ends with a consonant by permuting the letters of the word ‘HARSHITA’
A) 2340
B) 2700
C) 1800
D) 1980

Answer 1

According to given in the question the word is ‘HARSHITA’ for which we have to find number of words which begins with a vowel and ends with a consonant by permuting the letters of the word so first of all we have to count the number of the letters in the given word ‘HARSHITA’ and we also have to find the number of vowels and the consonant in the given word are as asked in the question that we have to find the number of words that starts with vowel and ends with consonant. After finding the number and letters of vowels and consonants for the given word now we can find the total number of words that can be started with the vowels which are (AAI) for the given word. But, as we know repetition is not allowed so we divide the number of letters that are repeated. Now, we will find the number of words that starts and ends with a vowel but, according to the given word it makes two cases. Now, to find the number of words that begins with a vowel and ends with a consonant by permuting the letters of the word ‘HARSHITA’ we would have to subtract the total number of words starts and end with a vowel with the total number of words starts and ends with a vowel.

Formula used: $n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times (n - 4) \times (n - 5).................... \times 2 \times 1$………………………(a)

Complete step-by-step answer:
Given word,
‘HARSHITA’
Step 1: first of all we will count the number of the letters in the given word ‘HARSHITA’ hence, there are total 8 letters (H, A, R, S, H, I, T, and A)
Step 2: Now, we will count the total number of vowels in the given word ‘HARSHITA’ hence, there are total 2 vowels (A, A, and I)
Step 3: Now, Now, we will count the total number of consonant in the given word ‘HARSHITA’ hence, there are total 5 consonants (H, H, R, S, and T)
Step 4: Now, we can find the total number of words which starts with the vowels are as given below. According to the given word ‘HARSHITA’ as asked in the question we have find the words that starts with vowels can be start with (A, and I) so we will take these two letters as a one letter now the total number of letters when these two letters (A, and I ) are taken as one letter or unit are 7 including other letters in the given word ‘HARSHITA’ and the number of letters that can be end with the consonant which are (H, H, R, S, and T) will also be 7 because the last letter is fixed for consonant.
Hence,
Total number of words starts with vowels:
$= \dfrac{{7!}}{{2!}} + \dfrac{{7!}}{{2! \times 2!}}$
On solving the expression obtained just above with the help of the formula (a) as mentioned in the solution hint.

$$= \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2!}}{{2!}} + \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2!}}{{2! \times 2!}} \\\ = 7 \times 6 \times 5 \times 4 \times 3 + 7 \times 6 \times 5 \times 2 \times 3 \\\ = 2520 + 1260 \\\ = 3780 \\\$$

Therefore,
Total number of words starts with vowels: $3780$
Step 5: Now, we will find the number of words that can be start and end with a vowel (there are two cases AA and IA) so,
$= \dfrac{{6!}}{{2!}} + 6!$……………………..(1)
In the expression(1) as given above we can see that we are going to find the number of words that starts and ends with a vowel hence, there will be only 6 letters left because two vowels (A, and I) are already fixed for the first and the last place.
On solving the expression (1) obtained above with the help of the formula (a) as mentioned in the solution hint.

$$= \dfrac{{6 \times 5 \times 4 \times 3 \times 2!}}{{2!}} + 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\\ = 6 \times 5 \times 4 \times 3 + 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\\ = 360 + 720 \\\ = 1080 \\\$$

Step 6: Now, to find the number of words that starts with vowels and with the consonant we will to subtract the total number of words that starts and ends with a vowel with the total number of words that starts and ends with a vowel.
$= 3780 - 1080 \\\ = 2700 \\\$
Hence, the number of words which begins with a vowel and ends with a consonant by permuting the letters of the word ‘HARSHITA’ are: $2700$

Option B is the correct answer.

Note: The number of permutations of n distinct objects is $n \times (n - 1) \times (n - 2) \times ...... \times 2 \times 1,$which number is called n factorial and can be written as $n!$
A factorial is when a number has a point after it so it represents all the positive integers leading up to that number then you multiply to solve the factorial.
Until it is said in the question that repetition is allowed we will consider that the repletion is not allowed in the given question.