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The number of words of four letters containing equal number of vowels and consonants, where repetition is allowed, is

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$105^{2}$

$210 \times 243$

$105 \times 243$

$150 \times 21^{2}$

answer is D.

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Detailed Solution

2 places for vowels which can be selected in $4 C_{2}$ ways this 2 places can be filled by 25 ways (repetition) allowed the remaining 2 places can be filled by $21^{2}$ ways.
Total no of ways $21^{2} \times 5^{2} \times 4 C_{2} = 150 \times 21^{2}$

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