$\begin{array}{1 1}(A)\;1240\\(B)\;1840\\(C)\;1820\\(D)\;2005\end{array} $

- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

The number of ways of choosing first couple is =$15C_1\times 15C_1$

$\Rightarrow (15)^2$

The number of ways of choosing second couple is =$14C_1\times 14C_1$

$\Rightarrow (14)^2$

Thus the number of ways of choosing the couple is =$15^2+14^2+13^2+.....+2^2+1^2$

$\Rightarrow \large\frac{15\times(15+1)[2(15+1)]}{6}$

$\Rightarrow 1240$

Hence (A) is the correct answer.

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