Answer
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Hint: Try to make functions from given sets.
We know, $A = \left\{ {1,2,3,4} \right\}$ and $B = \left\{ {a,b,c,d} \right\}$
A function from $A$ to $B$ is said to be bijection if it is one-one and onto. This means different elements of $A$ has different images in $B$.
Also, each element of $B$ has preimage in $A$.
Let ${f_1},{f_2},{f_3}$and${f_4}$are the functions from $A$ to $B$.
\[
{f_1} = \left\{ {\left( {1,a} \right),\left( {2,b} \right),\left( {3,c} \right),\left( {4,d} \right)} \right\} \\
{f_2} = \left\{ {\left( {1,b} \right),\left( {2,c} \right),\left( {3,d} \right),\left( {4,a} \right)} \right\} \\
{f_3} = \left\{ {\left( {1,c} \right),\left( {2,d} \right),\left( {3,a} \right),\left( {4,b} \right)} \right\} \\
{f_4} = \left\{ {\left( {1,d} \right),\left( {2,a} \right),\left( {3,b} \right),\left( {4,c} \right)} \right\} \\
\]
We can verify that${f_1},{f_2},{f_3}$and${f_4}$ are bijective from $A$ to $B$.
Now,
\[
{f_1}^{ - 1} = \left\{ {\left( {a,1} \right),\left( {b,2} \right),\left( {c,3} \right),\left( {d,4} \right)} \right\} \\
{f_2}^{ - 1} = \left\{ {\left( {b,1} \right),\left( {c,2} \right),\left( {d,3} \right),\left( {a,4} \right)} \right\} \\
{f_3}^{ - 1} = \left\{ {\left( {c,1} \right),\left( {d,2} \right),\left( {a,3} \right),\left( {b,4} \right)} \right\} \\
{f_4}^{ - 1} = \left\{ {\left( {d,1} \right),\left( {a,2} \right),\left( {b,3} \right),\left( {c,4} \right)} \right\} \\
\]
Note: A bijection, bijective function, one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set.
We know, $A = \left\{ {1,2,3,4} \right\}$ and $B = \left\{ {a,b,c,d} \right\}$
A function from $A$ to $B$ is said to be bijection if it is one-one and onto. This means different elements of $A$ has different images in $B$.
Also, each element of $B$ has preimage in $A$.
Let ${f_1},{f_2},{f_3}$and${f_4}$are the functions from $A$ to $B$.
\[
{f_1} = \left\{ {\left( {1,a} \right),\left( {2,b} \right),\left( {3,c} \right),\left( {4,d} \right)} \right\} \\
{f_2} = \left\{ {\left( {1,b} \right),\left( {2,c} \right),\left( {3,d} \right),\left( {4,a} \right)} \right\} \\
{f_3} = \left\{ {\left( {1,c} \right),\left( {2,d} \right),\left( {3,a} \right),\left( {4,b} \right)} \right\} \\
{f_4} = \left\{ {\left( {1,d} \right),\left( {2,a} \right),\left( {3,b} \right),\left( {4,c} \right)} \right\} \\
\]
We can verify that${f_1},{f_2},{f_3}$and${f_4}$ are bijective from $A$ to $B$.
Now,
\[
{f_1}^{ - 1} = \left\{ {\left( {a,1} \right),\left( {b,2} \right),\left( {c,3} \right),\left( {d,4} \right)} \right\} \\
{f_2}^{ - 1} = \left\{ {\left( {b,1} \right),\left( {c,2} \right),\left( {d,3} \right),\left( {a,4} \right)} \right\} \\
{f_3}^{ - 1} = \left\{ {\left( {c,1} \right),\left( {d,2} \right),\left( {a,3} \right),\left( {b,4} \right)} \right\} \\
{f_4}^{ - 1} = \left\{ {\left( {d,1} \right),\left( {a,2} \right),\left( {b,3} \right),\left( {c,4} \right)} \right\} \\
\]
Note: A bijection, bijective function, one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set.
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