Question

Find the number of words which begin with a vowel and ends with a consonant by permuting the letters of the word ‘HARSHITA’.

Answer 1

Hint:
Here, we need to find the number of words that begin with a vowel and end with a consonant using the letters of the word ‘HARSHITA’. We can solve this problem by taking four cases, and adding their results. The four cases will be the word starts with A and ends with H, the word starts with I and ends with H, the word starts with A and ends with a consonant other than H, and the word starts with I and ends with a consonant other than H. There are 8 spaces to be filled.

Complete step by step solution:
The number of letters in the word ‘HARSHITA’ are 8, where A and H come twice.
The number of permutations to arrange \[n\] letters is given by \[n!\], where no letter is repeated.
The number of permutations to arrange \[n\] letters is given by \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\], where a letter appears \[{r_1}\] times, another letter repeats \[{r_2}\], and so on.
We can find the answer using four cases.

Case 1: The first space has an A, and the eighth space has a H.
Since 1 A and 1 H are fixed at the two spaces, we have 6 remaining letters to be placed in 6 spaces.
The 6 remaining letters are R, S, H, I, T, A.
We observe that no letter is being repeated.
Therefore, the number of words that start with an A and ends with a H, is
\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\]

Case 2: The first space has an I, and the eighth space has a H.
Since 1 I and 1 H are fixed at the two spaces, we have 6 remaining letters to be placed in 6 spaces.
The 6 remaining letters are R, S, H, A, T, A.
We observe that the letter A is being repeated twice.
Therefore, the number of words that start with an I and ends with a H, is
\[\dfrac{{6!}}{{2!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 360\]

Case 3: The first space has an A, and the eighth space has a consonant other than H.
The letter A is fixed at the first place.
The eighth space has a consonant other than H, that is R, S, or T.
Therefore, there are 3 ways to fill the eighth space.
Now, we observe that the 6 remaining letters will have the letter H twice.
Therefore, the number of words that start with an A and ends with a consonant other than H, is
\[3 \times \dfrac{{6!}}{{2!}} = 3 \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 1080\]

Case 4: The first space has an I, and the eighth space has a consonant other than H.
The letter I is fixed in the first place.
The eighth space has a consonant other than H, that is R, S, or T.
Therefore, there are 3 ways to fill the eighth space.
Now, we observe that the 6 remaining letters will have the letters A and H twice.
Therefore, the number of words that start with an I and ends with a consonant other than H, is
\[3 \times \dfrac{{6!}}{{2!2!}} = 3 \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 540\]
Now, we will add the results of the four cases to get the number of words that start with a vowel and end with a consonant.
Number of words \[ = 720 + 360 + 1080 + 540 = 2700\]

Therefore, the number of words which begin with a vowel and end with a consonant by permuting the letters of the word ‘HARSHITA’ is 2700.

Note:
We can also solve this problem using the subtraction rule. We can find that the number of words that start with a vowel are
\[\begin{array}{l}\dfrac{{7!}}{{2!}} + \dfrac{{7!}}{{2!2!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} + \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}\\ = 2520 + 1260\\ = 3780\end{array}\]
Similarly, we can find that the number of words that start with a vowel and end with a vowel are
\[\begin{array}{l}\dfrac{{6!}}{{2!}} + 6! = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} + 6 \times 5 \times 4 \times 3 \times 2 \times 1\\ = 360 + 720\\ = 1080\end{array}\]
Therefore, using the subtraction rule, we get the number of words which begin with a vowel and ends with a consonant as \[3780 - 1080 = 2700\].