Question

The letters of the word AEIOU are permuted in all possible ways and the words thus formed are arranged as in a dictionary, the rank of the word UIOEA is:
(A) 118
(B) 116
(C) 114
(D) 120

Answer 1

Hint: We know the order of the letters that are used while positioning the words in the dictionary. The order is A, then E, then I, and then O. When we have n letters and we have to find the number of possible ways to make words, then the number of possible words is \[n!\] . First find the number of words starting with the letter A. We have to form 5 letter words and since the first letter is A, we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is \[4!\] . Similarly, find the numbers of words starting with the letter E, I, and O. Then, find the number of words having the first two letters as UA and UE. Now, we have the words with UIA and UIE as their first three letters. Now, the next 5 letter words are UIOAE and UIOEA. Now, find the position of the last word UIOEA.

Complete step by step solution:
According to the question, it is given that the word we have is “AEIOU”. In this word, we have the letters A, E, I, O, and U.
We know the order of the letters that are used while positioning the words in the dictionary. The order is A, then E, then I, then O, and then U.
In the beginning, we have all the words starting with the letter A.
We have to form 5 letter words and since the first letter is A, we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is \[4!\] .
The number of words starting with the letter A = \[4\times 3\times 2\times 1=24\] .
After all the words starting with letter A, we have the words starting with the letter E.
 We have to form 5 letter words and since the first letter is E, we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is \[4!\] .
The number of words starting with the letter E = \[4!=4\times 3\times 2\times 1=24\] .
After all the words starting with letter E, we have the words starting with the letter I.
We have to form 5 letter words and since the first letter is I, we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is \[4!\] .
The number of words starting with the letter I = \[4!=4\times 3\times 2\times 1=24\] .
After all the words starting with letter I, we have the words starting with the letter O.
We have to form 5 letter words and since the first letter is O, so we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is \[4!\] .
The number of words starting with the letter O = \[4!=4\times 3\times 2\times 1=24\] .
After all the words starting with the letter O, we have the words having UA as their first two letters.
We have to form 5 letter words and since the first two letters of the words are UA, we have 3 letters to be filled at the three positions. The number of ways that the three letters can be arranged is \[3!\] .
The number of words having SA as their first two letters = \[3!=3\times 2\times 1=6\] .
After the words having UA as their first two letters, we have the words whose first two letters are starting with the letter UE.
We have to form 5 letter words and since the first two letters of the words are UE, we have 3 letters to be filled at the three positions. The number of ways that the three letters can be arranged is \[3!\] .
The number of words having UE as their first two letters = \[3!=3\times 2\times 1=6\] .
After all the words having its first two letters as UE, we have the words having UIA as their first three letters.
We have to form 5 letter words and since the first three letters of the words are UIA, so we have 2 letters to be filled at the three positions. The number of ways that the three letters can be arranged is \[2!\] .
The number of words having UIA as their first three letters = \[2!=2\times 1=2\] .
After all the words having its first three letters as UIA, we have the words having UIE as their first three letters.
We have to form 5 letter words and since the first three letters of the words are UIE, so we have 2 letters to be filled at the three positions. The number of ways that the three letters can be arranged is \[2!\] .
The number of words having UIA as their first three letters = \[2!=2\times 1=2\] .
Now, the next 5 letter words are UIOAE and UIOEA.
Therefore, the position of the word UIOEA will be,
\[\begin{align}
  & 24+24+24+24+6+6+2+2+2 \\
 & =96+12+6 \\
 & =114 \\
\end{align}\]
Hence, the correct option is (C).

Note: In this question, we might think that UIOEA is going to be our last word and we might get the position of this word as \[5!\] which is equal to 120. This is wrong. Our last word is UOIEA and we have to find the position of the word UIOEA. So, the position of UIEOA cannot be equal to 120.