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The number of 4 letter permutations formed with 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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Verified Answer
The correct answer is:
$3^6 \times 70$
The permutation of 4 letters such that voweI and consonants are equal
$$
\begin{aligned}
{ }^{21} C_1 & \times{ }^5 C_1 \times \frac{4 !}{2 ! 2 !}+{ }^{21} C_2 \times{ }^5 C_2 \times 4 ! \\
& =21 \times 5 \times 6+21 \times 10 \times 5 \times 2 \times 24 \\
& =630+50400 \\
& =51030 \\
& =729 \times 70=3^6 \times 70
\end{aligned}
$$
$$
\begin{aligned}
{ }^{21} C_1 & \times{ }^5 C_1 \times \frac{4 !}{2 ! 2 !}+{ }^{21} C_2 \times{ }^5 C_2 \times 4 ! \\
& =21 \times 5 \times 6+21 \times 10 \times 5 \times 2 \times 24 \\
& =630+50400 \\
& =51030 \\
& =729 \times 70=3^6 \times 70
\end{aligned}
$$
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