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Let $\mathbf{x}=\{\mathbf{1}, \mathbf{2}, \mathbf{3}, \mathbf{4}, 5\} .$ The number of different ordered pairs. $(\mathbf{y}, \mathbf{z})$ that can be formed such that $Y \in X, Z \in X$ and $y \cap z$ is empty, is

1) $5^{2}$
2) $3^{5}$
3) $2^{3}$
4) $5^{3}$
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(2)

Explanation

$x_{1} \in y, x_{1} \in z$
$x_{1} \notin y, x_{1} \in z$
$x_{1} \in y, x_{1} \notin z$
$y \cap z=\phi$
$x_{1} \notin y_{1} x_{1} \notin z$
each $x_{1} \in X$ having 3 chances
$n(y \cap z)=3^{5}$

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