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Question: Find the number of words which begin with a vowel and ends with a consonant by permuting the letters...

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

Explanation

Solution

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 nn letters is given by n!n!, where no letter is repeated.
The number of permutations to arrange nn letters is given by n!r1!r2!rn!\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}, where a letter appears r1{r_1} times, another letter repeats r2{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!=6×5×4×3×2×1=7206! = 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
6!2!=6×5×4×3×2×12×1=360\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
3×6!2!=3×6×5×4×3×2×12×1=10803 \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
3×6!2!2!=3×6×5×4×3×2×12×1×2×1=5403 \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 = 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
7!2!+7!2!2!=7×6×5×4×3×2×12×1+7×6×5×4×3×2×12×1×2×1 =2520+1260 =3780\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
6!2!+6!=6×5×4×3×2×12×1+6×5×4×3×2×1 =360+720 =1080\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 37801080=27003780 - 1080 = 2700.