A is a set having 6 distinct elements. The number of distinct functions from A to A which are not bijections are?
A. $6! - 6$
B. ${6^6} - 6$
C. ${6^6} - 6!$
D. $6!$
Answer
364.2k+ views
Hint: - First we will find the total number of functions from A to A then we will subtract the number of functions which are bijections.
We know that if a set $S$ has $n$ distinct elements then the number of functions from $S$ to $S$ is given by${n^n}$. (Formulae)
Since, A is the set having 6 distinct elements.
Then total number of functions from A to A are = ${6^6}$
As we know that the functions which are one-one will be onto as well, as the mapping is from the same set to the same set.
We know that if a set $S$ has $n$distinct elements then the number of one-one and onto functions from $S$ to $S$ is given by $n!$. (Formulae)
So, the number of functions which are both one-one and onto (i.e. bijective) are = $6!$
Hence, the total number of distinct functions from A to A which is not bijections is:
Total number of functions $-$ Number of bijective functions
$ = {6^6} - 6!$
Hence, option C is correct.
Note: - A bijective 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. There is no method to find the number of non-bijective functions, so we have counted the total number of functions and subtracted the number of bijective functions.
We know that if a set $S$ has $n$ distinct elements then the number of functions from $S$ to $S$ is given by${n^n}$. (Formulae)
Since, A is the set having 6 distinct elements.
Then total number of functions from A to A are = ${6^6}$
As we know that the functions which are one-one will be onto as well, as the mapping is from the same set to the same set.
We know that if a set $S$ has $n$distinct elements then the number of one-one and onto functions from $S$ to $S$ is given by $n!$. (Formulae)
So, the number of functions which are both one-one and onto (i.e. bijective) are = $6!$
Hence, the total number of distinct functions from A to A which is not bijections is:
Total number of functions $-$ Number of bijective functions
$ = {6^6} - 6!$
Hence, option C is correct.
Note: - A bijective 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. There is no method to find the number of non-bijective functions, so we have counted the total number of functions and subtracted the number of bijective functions.
Last updated date: 24th Sep 2023
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Total views: 364.2k
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