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+23 votes
How many onto (or surjective) functions are there from an $n$-element $(n ≥ 2)$ set to a $2$-element set?

  1. $ 2^{n}$
  2. $2^{n} – 1$
  3. $2^{n} – 2$
  4. $2(2^{n} – 2)$
asked in Set Theory & Algebra by Boss (16k points) | 2.2k views

4 Answers

+26 votes
Best answer

No. onto (or surjective) functions are there from an $n$-element $(n\geq 2)$ set to a $2$-element set $=$ Total No of functions $-$ (No of functions with $1$ element from RHS not mapped) $+$ (No of functions with $2$ element from RHS not mapped)$\ldots$(So on Using Inclusion Excusion principle $= 2^{n}$ (Total no of functions ) $-$ $2 * 1^{n}$ ( No of functions in which one element is excluded) $+ 0$ (No element in RHS is selected) $= 2^{n} -2.$

Hence Ans is (C).


answered by Boss (41k points)
edited by
$\sum_{i=0}^{n-1} (-1)^i{n \choose i}(n-i)^m $ =
${n \choose 0}n^m - {n \choose 1}(n-1)^m + {n \choose 2}(n-2)^m - \cdots \pm {n \choose n-2}2^m \mp {n \choose n-1}1^m $
@akash can you please show how to do this question with that alternate approach of distribution ?
Total number of functions $f: A\rightarrow B$ possible where |A| = n and |B|= 2 is $2^n$, out of which only 2 functions are there that are not onto-
1. When all n elements of $A$ mapped to $1^{st}$ element of $ B$.
2. When all n elements of $A$ mapped to $2^{nd}$ element of $ B$.

So $2^n - 2$ onto functions are possible.

Onto functions

can u pls tell me how u took that probability using bernouilli

i  mean how u did the 2nd last step @Prateek Raghuvanshi

see this please if it is correct!!

# of onto fn= total fn - into fn

=2n- {atleast one of rhs is not mapped}

=2n - {1- all elements in rhs are mapped}

now am not able to solve it further;

the way u did is also good, but i was thinking if the elements in set B are large,then we cant chk for every case right, we have to do it like this.


can u pls tell me how to solve it further. It would be  a great help!


Gate Fever check this out-

+21 votes

$2^{n} - 2$

in words (Total functions $-$ 2 functions where all elements maps exactly one element).

answered by Boss (14.4k points)
edited by
+7 votes

Consider |A| = a, |B| = b, then Number of relations from A to B is $2^{a \times b }$, as each element is A can map to every element is B, and each mapping may exist or not.

Now in case of a function its one - one or many to one. Hence total functions possible is $b^{a}$, as each value in A can map to only one value in B, and this is applicable 'a' times.

With this information in mind, number of functions from A to B in this case is $2^n$ But two of the functions are not onto i.e. the ones where all map to only 1st or 2nd element in B.

Hence answer is C

answered by Active (3.5k points)
+4 votes

$Onto function :\space$ $x$  $\longrightarrow$ $y$

$x\space is\space preimage\space of\space y\space ,\space y\space is\space image\space of\space x\space$

$\textbf {Step 1:}Find\space non\space onto\space function \space is\space relatively\space easythan\space find\space onto\space function\space so\space -\space$

$non\space onto\space function\space =\space Either\space a\space not\space mapped\space OR\space B\space not\space mapped\space OR\space C\space not\space mapped\space$

${So\space we\space need\space to\space find\space}$ $|$$S_a$ $\cup$ $S_b$ $\cup$  $S_c$$|$ $=\space ?$ 

$\textbf{Step 2:}\space According\space to\space \textbf{Principal of Mutual Inclusion and Exclusion}$

$|$ $S_a$ $\cup$  $S_b$ $|$ = $|$ $S_a$ $|$ $+$  $|$ $S_b$ $|$ $-$ $|$ $S_a$ $\cap$ $S_b$ $|$

$|$ $S_a$ $|$ $=$ $It$ $means\space a\space not\space mapped$ $=$ $1^n$ 

$|$ $S_b$ $|$ $=$ $It$ $means\space b\space not\space mapped$ $=$ $1^n$ 

$if\space both\space not\space mapped$ $=$ $0^n$

$|$ $S_a$ $\cup$  $S_b$ $|$ = $1\space +\space 1\space -\space 0\space =\space 2$

$\textbf{Step 3:}\space So\space Onto\space functions\space =\space Total\space functions\space -\space non\space onto\space functions $

$And\space as\space we\space know\space total\space functions\space =\space 2^n$

$So\space onto\space functions\space are\space =\space 2^n - 2$

$Hence\space Option\space C\space is\space right\space one$

answered by Boss (10.5k points)
edited by

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