$\begin{array}{1 1}(A)\;455\\(B)\;1575\\(C)\;1120\\(D)\;2030\end{array} $

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

Number are either all even or one even and other two odd.

Required number of ways =$15C_3+15C_1\times 15C_2$

$15C_3=\large\frac{15!}{3!12!}=\frac{15\times 14\times 13\times 12!}{3\times 2\times 12!}$

$\Rightarrow 455$

$15C_1=\large\frac{15!}{1!\times 14!}$

$\Rightarrow 15$

$15C_2=\large\frac{15!}{2!\times 13!}$

$\Rightarrow 105$

$15C_3+15C_1\times 15C_2=455+15\times 105$

$\Rightarrow 455+1575$

$\Rightarrow 2030$

Hence (D) is the correct answer.

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