Question
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’.
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 n letters is given by n!, where no letter is repeated.
The number of permutations to arrange n letters is given by r1!r2!…rn!n!, where a letter appears r1 times, another letter repeats r2, 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=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
2!6!=2×16×5×4×3×2×1=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
3×2!6!=3×2×16×5×4×3×2×1=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
3×2!2!6!=3×2×1×2×16×5×4×3×2×1=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
2!7!+2!2!7!=2×17×6×5×4×3×2×1+2×1×2×17×6×5×4×3×2×1 =2520+1260 =3780
Similarly, we can find that the number of words that start with a vowel and end with a vowel are
2!6!+6!=2×16×5×4×3×2×1+6×5×4×3×2×1 =360+720 =1080
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.