Question

The number of different words of three letters which can be formed from the word "PROPOSAL", if a vowel is always in the middle are:
A) 53
B) 52
C) 63
D) 32

Answer 1

Here we will first count the number of letters present in the given word and then we will count the number of vowels and number of consonants present in the word. Then we will take different cases considering the vowels. Then we will find the sum of the number of ways in different cases and solve it further until we will get our required answer.

Complete Step by Step Solution:
Here we need to find the number of different three letters words that can be formed using the letters in the given word i.e. "PROPOSAL".
We can see that there are a total of eight letters in the word "PROPOSAL".
There are three vowels i.e. ‘O’, ‘O’, ‘A’ and there are five consonants i.e. ‘P’, ‘P’, ‘R’, ‘S’, ‘L’.
In first case, we will consider the vowel ‘O’ in the middle i.e. $\underline {} \underline {\rm{O}} \underline {}$
The letters left are ‘P’, ‘P’, ‘R’, ‘S’, ‘L’, ‘O’, ‘A’.
As we can see that the letter ‘P’ has occurred two times, we will consider it as one letter. Number of letters left is 6.
Total number of ways to arrange these letters in the first and the third position $= {}^6{P_2}$
Simplifying using the formula${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= \dfrac{{6!}}{{\left( {6 - 2} \right)!}}$
Subtracting the terms in the denominator, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= \dfrac{{6!}}{{4!}}$
Computing the factorial, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= \dfrac{{6 \times 5 \times 4!}}{{4!}}$
Dividing the terms, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= 6 \times 5 = 30$
There will be one more word possible which we have not included here i.e. $\underline P \underline {\rm{O}} \underline P$
Number of words formed when the vowel ‘O’ is in the middle $= 30 + 1 = 31$ ………… $\left( 1 \right)$

In second case, we will consider the vowel ‘A’ in the middle i.e. $\underline {} \underline A \underline {}$
The letters left are ‘P’, ‘P’, ‘R’, ‘S’, ‘L’, ‘O’, ‘O’.
As we can see that the letter ‘P’ and the letter ‘O’ has occurred two times, we will consider it as one letter. Number of letters left is 5.
Total number of ways to arrange these letters in the first and the third position $= {}^5{P_2}$
Simplifying using the formula${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= \dfrac{{5!}}{{\left( {5 - 2} \right)!}}$
Subtracting the terms in the denominator, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= \dfrac{{5!}}{{3!}}$
Computing the factorial, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= \dfrac{{5 \times 4 \times 3!}}{{3!}}$
Dividing the terms, we get
$\Rightarrow$ Total number of ways to arrange these letters in the first and the third position $= 5 \times 4 = 20$
There will be two more words possible which we have not included here i.e. $\underline O \underline A \underline O$ and $\underline P \underline A \underline P$
Number of words formed when the vowel ‘A’ is in the middle $= 20 + 1 + 1 = 22$ ………… $\left( 2 \right)$
Hence, the total number of words that can be formed from the given word $= 22 + 31 = 53$

Thus, the correct option is option A.

Note:
Here we need to know the basic formulas of permutation and combination. Permutation is used when we have to find the possible elements but the combination is used when we need to find the number of ways to select a number from the collection. In permutation the order of arrangement matters, whereas in combination, the order of selection, does not matter.
Here, we have to add the possible number of ways to get the required value. If we don't add the possible ways we will not get the correct answer.