If $n (A) = m, m > 0$ , then number of symmetric relations from $A$ to $A$ is

2^{m^{2} - m}

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2^{\frac{m^{2} - m}{2}}

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2^{m^{2} + m}

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2^{\frac{m^{2} + m}{2}}

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Solution

The correct option is D $2^{\frac{m^{2} + m}{2}}$
Let $A = {a_{1}, a_{2}, a_{3}, \dots ., a_{m}}$
$\Rightarrow A \times A = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (a_{1}, a_{1}), (a_{1}, a_{2}), \dots, (a_{1}, a_{m}) (a_{2}, a_{1}), (a_{2}, a_{2}), \dots, (a_{2}, a_{m}) . . . . (a_{m}, a_{1}), (a_{m}, a_{2}), \dots, (a_{m}, a_{m}) \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$
$n (A \times A) = m^{2}$
Let us Consider elements in $A \times A$ as
$1)$ Diagonal elements $D_{1} = {(a_{1}, a_{1}), (a_{2}, a_{2}) . . . (a_{m}, a_{m})}$
$2)$ Non diagonal elements $D_{2} = {(a_{1}, a_{2}), (a_{2}, a_{1}), (a_{1}, a_{3}), (a_{3}, a_{1}) . . . (a_{m - 1}, a_{m}), (a_{m}, a_{m - 1})}$
$n (D_{1}) = m, n (D_{2}) = m^{2} - m$
Now for relation to be symmetric
$1)$ there are two choices for each of $m$ diagonal elements
$2)$ For $m^{2} - m$ non diagonal elements we can form pairs of $(a_{1}, a_{2}), (a_{2}, a_{1}); (a_{1}, a_{3}) (a_{3}, a_{1}); (a_{m - 1}, a_{m}) (a_{m}, a_{m - 1});$
for each pair there are two choices, either it can be included or excluded to make relation symmetric.
$∴$ From principle of counting,number of symmetric relations
$= 2 \times 2 \times . . .2 (m times) \times 2 \times 2 \times . . .2 (\frac{m^{2} - m}{2} times) = 2^{\frac{m^{2} + m}{2}}$

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