$\begin{array}{1 1}(A)\;50000\\(B)\;50100\\(C)\;50300\\(D)\;50400\end{array} $

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$
- $n!=n(n-1)(n-2)(n-3).......(3)(2)(1)$

2 vowels can be chosen in $5C_2$ ways

2 consonants can be chosen in $21C_2$ ways

4 letters can be arranged in 4! ways

$\therefore$ The number of words consisting of 2 vowels and 2 consonants

$\Rightarrow 5C_2\times {21}C_2\times 4!$

$\Rightarrow \large\frac{5\times 4}{1\times 2}\times \frac{21\times 20}{1\times 2}$$ \times 24$

$\Rightarrow 10\times 210\times 24$

$\Rightarrow 50400$

Hence (D) is the correct answer.

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