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 $${\displaystyle \frac{n!}{r_1!r_2!\dots r_n!}}$$, 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
$${\displaystyle \frac{6!}{2!}}={\displaystyle \frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1}}=360$$

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
$$3\times {\displaystyle \frac{6!}{2!}}=3\times {\displaystyle \frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1}}=1080$$

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
$$3\times {\displaystyle \frac{6!}{2!2!}}=3\times {\displaystyle \frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}=540$$
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
$$\begin{array}{l}{\displaystyle \frac{7!}{2!}}+{\displaystyle \frac{7!}{2!2!}}={\displaystyle \frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 1}}+{\displaystyle \frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}\\ =2520+1260\\ =3780\end{array}$$
Similarly, we can find that the number of words that start with a vowel and end with a vowel are
$$\begin{array}{l}{\displaystyle \frac{6!}{2!}}+6!={\displaystyle \frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1}}+6\times 5\times 4\times 3\times 2\times 1\\ =360+720\\ =1080\end{array}$$
Therefore, using the subtraction rule, we get the number of words which begin with a vowel and ends with a consonant as $$3780-1080=2700$$.