Question

Consider the name of a baby girl ‘HARSHITA’. New names are formed (with or without meaning) by using all the letters of the given word.
On the basis of the above information, answer the following question. Number of words in which vowels occupy even places.

Answer 1

Hint: First of all we will count the number of words in the name of the baby girl “HARSHITA” and also expected to place vowels at even places so it will second, fourth, sixth and eighth letter’s position and will use the combinations formula to get the required value.

Complete step-by-step answer:
Given: The name of a baby girl ‘HARSHITA’ and here we are expected to place vowels at even places
HARSHITA is the eight lettered word and we have second, fourth, sixth and eighth positions for the vowels.
Out of a total of eight letters, three letters are vowels.
So, the possible ways of combinations are given by $ {}^4{C_3} $ ways. ….. (A)
Now, the arrangement of three vowels can be possibly done in
 $ \dfrac{{3!}}{{2!}} $ ways …..(B)
Now, the arrangement of the consonants can be done possibly in
 $ \dfrac{{5!}}{{2!}} $ ways …..(C)
Now, the total number of possible ways becomes
 $ = {}^4{C_3} \times \dfrac{{3!}}{{2!}} \times \dfrac{{5!}}{{2!}} $
Simplify the above expression –
Possible ways $ = \dfrac{{4 \times 3 \times 5 \times 4 \times 3 \times 2}}{2} $
Common terms from the numerator and the denominator cancel each other.
Possible ways $ = 4 \times 3 \times 5 \times 4 \times 3 $
Find the product of the above terms –
Possible ways $ = 720 $ ways
Hence, number of words in which vowels occupy even places are $ 720 $ ways
So, the correct answer is “ $ 720 $ ways”.

Note: First of all be clear about the difference between vowels and consonants. We have five vowels in English alphabets a, e, I, o, u and the rest of the alphabets are consonants. Use the combinations and permutations formula accordingly and then wisely simplify for the required value. Even numbers end with $ 0,2,4,6,8 $