Question

Find the number of bijective functions from set A to itself when A contains 106 elements.
$\left( A \right)$ 106
$\left( B \right)$ ${\left( {106} \right)^2}$
$\left( C \right)$ 106!
$\left( D \right)$ ${2^{106}}$

Answer 1

Hint – In this particular type of question use the concept that the bijective function of any set from itself is all of the ordered pairs and these ordered pairs are called as bijective functions, so use this concept to reach the solution of the question.

Complete step-by-step answer:
Bijection function of the given set is all of the possible ordered pairs, and these ordered pairs are called as bijective functions
Consider a set S which has 3 elements {a, b, c} so all of the ordered pairs for this set to itself i.e. S to S are (a, b), (b, c), (a, c), (b, a), (c, b), and (c, a).
So there are 6 ordered pairs i.e. 6 bijective functions which is equivalent to (3!).
So as we see that in the set A there are 3 elements so the total bijective functions to itself are (3!).
Now if there are n elements in any set so the number of ordered pairs are (n!). So, the number of bijective functions to itself are (n!).
Now it is given that in set A there are 106 elements.
Now we have to find out the number of bijective functions from set A to itself (i.e. A to A)
So from the above information the number of bijective functions to itself (i.e. A to A) is 106!
So this is the required answer.
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that if there are n elements in any set than all of the ordered pairs of this set to itself is (n!) and these all sets are called as bijective functions so in the given question, n = 106 so all the bijective functions are 106!