Question

The different words beginning and ending with a vowel that can be made with all the letters of the word EQUATION is:
a. 14400
b. 4320
c. 864
d. 1440

Answer 1

Hint: We start with checking initially how many vowels and consonants are here in this given word. Then we fix the letters in the starting and the ending. Then we rearrange the remaining letters among themselves to find our needed answer.

Complete step-by-step answer:
We have our word as, “EQUATION” where we have 8 letters. And we have 3 consonants which are Q, T and N and 5 vowels which are E, U, I, A, O.
Now to arrange in a way that vowels occupy first and last place and it can be done in \[{}^5{P_2}\] ways as we are choosing 2 places from 5 letters. We will now be left with 3 vowels and 3 consonants. So, we have 6 letters getting arranged with themselves which can be arranged in \[6!\;\] ways.
Hence the number of words under given condition is,
 \[{}^5{P_2} \times 6!\]
Using \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] , we get,
 \[ = \dfrac{{5!}}{{(5 - 2)!}} \times 6!\]
Now using \[n! = n(n - 1)!\] , we get,
 \[ = \dfrac{{5 \times 4 \times 3!}}{{3!}} \times 6!\]
Now on canceling we get,
 \[ = 5 \times 4 \times 6!\]
Now as \[6! = 720\] , we get,
 \[ = 20 \times 720\]
On multiplying we get,
 \[ = 14400\] ways,
So, The different words beginning and ending with a vowel that can be made with all the letters of the word EQUATION is 14400.
Hence, option (a) is correct.

Note: We have to keep in mind that when we use permutation and when combination while dealing with a probability problem. Permutation is used for lists (order matters) and Combination for groups (order doesn’t matter).