
The different words beginning and ending with a vowel that can be made with all the letters of the word ‘EQUATION’ is:
$\eqalign{
& A)\,\,\,14400 \cr
& B)\,\,\,4320 \cr
& C)\,\,\,864 \cr
& D)\,\,1440 \cr} $
Answer
586.2k+ views
Hint:
First, we have to calculate the number of words formed beginning and ending with a vowel. In between them,we have to look at the total possibilities so that the above condition will be satisfied.
Complete step by step solution:
Step1: The word ‘EQUATION’ has five vowels: ‘A’ ,’E’, ’I’, ’O’, ’U’.
Step2: To form a word starting with a vowel,there are 5 possibilities i.e. we can choose any single vowel among five vowels.
Step3: Again for each word starting with a vowel, can be ended with a vowel in 4 number of ways i.e. the remaining 4 vowels can be chosen.
Step4: There are 6 numbers of places in between the above two vowels in each word as the total number of letters or total number of places of the word is 8.
These 6 places in the above form each word can be filled up with 6! ways.
Step5: Hence total number of words
$\eqalign{
& = 5 \times 4 \times 6! \cr
& = 20 \times 720 \cr
& = 1440 \cr} $
Therefore, the different words beginning and ending with a vowel that can be made with all the letters of the word ‘EQUATION’ is 1440.
So,the option A) is correct here.
Note:
Here we use the formula, n number of places can be filled up with n number of distinct letters in n! ways.$Here\,\,n! = n(n - 1)(n - 2)(n - 3)...3\cdot 2 \cdot 1$.
First, we have to calculate the number of words formed beginning and ending with a vowel. In between them,we have to look at the total possibilities so that the above condition will be satisfied.
Complete step by step solution:
Step1: The word ‘EQUATION’ has five vowels: ‘A’ ,’E’, ’I’, ’O’, ’U’.
Step2: To form a word starting with a vowel,there are 5 possibilities i.e. we can choose any single vowel among five vowels.
Step3: Again for each word starting with a vowel, can be ended with a vowel in 4 number of ways i.e. the remaining 4 vowels can be chosen.
Step4: There are 6 numbers of places in between the above two vowels in each word as the total number of letters or total number of places of the word is 8.
These 6 places in the above form each word can be filled up with 6! ways.
Step5: Hence total number of words
$\eqalign{
& = 5 \times 4 \times 6! \cr
& = 20 \times 720 \cr
& = 1440 \cr} $
Therefore, the different words beginning and ending with a vowel that can be made with all the letters of the word ‘EQUATION’ is 1440.
So,the option A) is correct here.
Note:
Here we use the formula, n number of places can be filled up with n number of distinct letters in n! ways.$Here\,\,n! = n(n - 1)(n - 2)(n - 3)...3\cdot 2 \cdot 1$.
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