1.5k views

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. $m*n$
edited | 1.5k views
0
No of functions: $(size of set 2)^{size of set 1}$

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.
selected
0
+3

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.

$m$ 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 $B$

edited
–1 vote
m--->n

ans: C

1
2