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(C) We need to form a $4$-letter word with $2$ distinct vowels and $2$ distinct consonants,where repetition is allowed. Let $V$ be the set of $5$ vowe

Vedclass · Accepted Answer

$3^6 	imes 70$

Vedclass · Answer

(C) We need to form a $4$-letter word with $2$ distinct vowels and $2$ distinct consonants,where repetition is allowed. Let $V$ be the set of $5$ vowels and $C$ be the set of $21$ consonants. Case $1$: We choose $1$ vowel and $1$ consonant,and each is repeated once. The number of ways to choose $1$ vowel and $1$ consonant is $\binom{5}{1} 	imes \binom{21}{1} = 5 	imes 21 = 105$. The number of arrangements of these $4$ letters (e.g.,$V_1 V_1 C_1 C_1$) is $\frac{4!}{2!2!} = 6$. Total for Case $1 = 105 	imes 6 = 630$. Case $2$: We choose $2$ distinct vowels and $2$ distinct consonants. The number of ways to choose $2$ vowels and $2$ consonants is $\binom{5}{2} 	imes \binom{21}{2} = 10 	imes 210 = 2100$. The number of arrangements of these $4$ distinct letters is $4! = 24$. Total for Case $2 = 2100 	imes 24 = 50400$. Total permutations $= 630 + 50400 = 51030$. $51030 = 729 	imes 70 = 3^6 	imes 70$.

The number of $4$-letter permutations formed using the English alphabet such that the number of distinct vowels is equal to the number of distinct consonants,when repetition is allowed,is

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