$\begin{array}{1 1}(A)\;1000\\(B)\;1430\\(C)\;1440\\(D)\;1560\end{array} $

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- $n!=n(n-1)(n-2)......(3)(2)(1)$

The word EQUATION consists of 5 vowels and 3 consonants.

5 words can be arranged in 5!

$5!=5\times 4\times 3\times 2\times 1$

$\;\;\;\;=120$ ways

3 consonants can be arranged in 3!

$3!=3\times 2\times 1$

$\;\;\;\;=6$ ways

The two block of vowels and consonants can be arranged in 2! ways

$2!=2\times 1$

$\;\;\;\;=2$ ways

$\therefore$ The number of words which can be formed with letters of the word EQUATION so that vowels and consonants occur together

$\Rightarrow 120\times 6\times 2$

$\Rightarrow 1440$

Hence (C) is the correct answer.

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