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+13 votes

The number of functions from an $m$ element set to an $n$ element set is

  1. $m + n$
  2. $m^n$
  3. $n^m$
  4. $m*n$
asked in Set Theory & Algebra by Veteran (52.1k points)
edited by | 1.5k views
No of functions: $(size of set 2)^{size of set 1}$

3 Answers

+25 votes
Best answer
No. of functions from an $m$ element set to an $n$ element set is $n^m$ as for each of the $m$ element, we have $n$ choices to map to, giving $\underbrace{n \times n \times \dots n}_{m \text{ times}} = n^m$.

PS: Each element of the domain set in a function must be mapped to some element of the co-domain set.
answered by Veteran (59.9k points)
selected by
It should be m^n..Please clarify

Let set A contains m element and set B contains n elements and we have to map from A to B. Then for every elements of A we have n choices to map to, and think recursively that 1st element we have n choice,  again for the 2nd element of A we have n choice and if we continue like this then every element of A has n choices. So n*n*n*n*n*n*..........upto m times will generate n^m. Like we write 2*2*2*2 = 2^4. i hope this clear your doubt.

+4 votes

$m$ which is domain and $n$ which is co-domain ,so number of functions = $co-domain^{domain }$

so $n^{m}$

$OR$ In another way

Every element of Set with m elements have $n$ options so for $1\rightarrow n \space\text{for }2\rightarrow n$ options ,

Similiarly for $m\rightarrow n$ options so

$n\times n \times n \space ...\text{m times} = n^m$

So option $C$ Not $B$

answered by Boss (10.6k points)
edited by
–1 vote

ans: C
answered by Loyal (7.1k points)

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