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: In this question we use the theory of permutation and combination. So, before solving this question we need to first recall the basics of this chapter. For example, if we need to select two alphabets out of four alphabets. Then in this case, this can be done in ${}^{\text{4}}{{\text{C}}_2}$ =6 ways.

Complete step-by-step answer:
Formula used- ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
Let us first select two places for vowels, which can be selected from 4 places in ${}^{\text{4}}{{\text{C}}_2}$ ways.
Now this place can be filled by vowels in
5×5=25 ways as repetition is allowed.
The remaining two places can be filled by consonants in 21×21 ways.
Then the total number of words is ${}^{\text{4}}{{\text{C}}_2}$$ \times 25 \times {21^2} = 150 \times {21^2}$
Therefore, the number of words of four letters containing equal numbers of vowels and consonants, where repetition is allowed, is $150 \times {21^2}$.
Thus, option (D) is the correct answer.

Note: A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. For example, suppose we have a set of three letters: A, B, and C. Each possible selection would be an example of a combination.