Question

The number of words of four letters containing equal number of vowels and consonants, where repetition is allowed, is
A.\[{105^2}\]
B.\[210 \times 243\]
C.\[150 \times 243\]
D.\[150 \times {21^2}\]

Answer 1

Hint: Here, we will find the number of ways of selecting two letters from the 4 lettered words such that those two letters contain vowels. Then, we will find the number of ways by which each letter can be filled by vowels. Similarly, we will find the number of ways by which the remaining two letters can be filled by consonants. Hence, multiplying all the ways together we will get the required number of words containing equal numbers of vowels and consonants.

Formula Used:
We will use the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], where \[n\] is the total number of letters in the word and \[r\] is the number of letters where vowels can be placed.

Complete step-by-step answer:
Now, first of all, with the help of combinations, we will select the 2 places out of the four places in the word where vowels can be placed.
Hence, substituting \[n = 4\] and \[r = 2\] in the formula of combinations \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get,
The number of places where the vowels can be placed \[ = {}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow \] The number of places where the vowels can be placed \[ = \dfrac{{4!}}{{2!2!}}\]
Computing the factorial, we get
\[ \Rightarrow \] The number of places where the vowels can be placed \[ = \dfrac{{4 \times 3}}{2}\]
Simplifying the expression, we get
\[ \Rightarrow \] The number of places where the vowels can be placed \[ = 2 \times 3 = 6\]……………\[\left( 1 \right)\]
Now, we know that the total numbers of vowels are 5.
As repetition is allowed, therefore, the number of ways of filling the first place of vowels is 5 and the number of ways of filling the second place of vowels is also 5.
Hence, the total number of ways of filling the 2 letters which are vowels are: \[{}^4{C_2} \times 5 \times 5\]……………\[\left( 2 \right)\]
Now, similarly, the remaining two places can be filled by the 21 consonants.
Since, repetition is allowed, the first place among the two remaining places can be filled by 21 consonants and the second place among the two remaining places can also be filled by 21 consonants only.
Hence, the total number of ways of filling the remaining two places\[ = 21 \times 21\]…………………………\[\left( 3 \right)\]
Hence, from equation \[\left( 2 \right)\] and equation \[\left( 3 \right)\], we get
Required number of words \[ = {}^4{C_2} \times 5 \times 5 \times 21 \times 21\]
Substituting the value of \[{}^4{C_2} = 6\] from equation \[\left( 1 \right)\], we get
\[ \Rightarrow \] Required number of words \[ = 6 \times 25 \times {21^2}\]
\[ \Rightarrow \] Required number of words \[ = 150 \times {21^2}\]
Therefore, The number of words of four letters containing equal number of vowels and consonants, where repetition is allowed is \[150 \times {21^2}\]
Hence, option D is the correct answer.

Note: To solve this question, we should know the difference between permutations and combinations. Permutation is an act of arranging the numbers or elements where the order of arrangement matters. Also, Combination is a method of selecting a group of numbers or elements in any order. In the case of combinations, the selection matters the most and not the order in which we have selected.