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Let $A$ be a finite set having $x$ elements and let $B$ be a finite set having $y$ elements. What is the number of distinct functions mapping $B$ into $A$.

1. $x^y$
2. $2^{(x+y)}$
3. $y^x$
4. $y! / (y-x)!$

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Set A have x elements and B set have y elements. Each elements in B has x choices to be mapped to and being a function it must map to some element.Since each element has exactly x choices,

The total number of functions from B to A =$\underbrace{x\times x\times x\times\cdots \times x }_{y \text{ times}} \\ = x^y.$

Hence, Option(A) xis the correct choice.

For example we can consider $A = \{1\}$ and $B=\{1, 2\}$. Now, only possible function from $B \to A$ is

1. $\{ (1,1), (2,1)\}$
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i think it should be Y..

plz check....

2 *2*2 =            23 = 8                 or             3= 9

which one is right ?

yes, i got it...
Don't know whether that was changed or not, but anyway A is the correct answer.

I don't think this Answer need an Explanation

by
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function mapping is B into A Not A into B. then how answer could be x^y.....I think i am confused please clear it .

yes,Answer should be xy . Your concept is also Right but you did calculation mistake.You said

"set B in which we have y elements has x no of choices either to choose a particular element from set A or not and set A ."

So, it becomes $\underbrace{x \times x \times \dots x}_{y\text{ times}} = x^y.$

ok what is 2*2*2 = ?

23  or 3 2  ?

let A ={1,2,3,4}

B={5,6}

for F:B->A, B IS THE DOMAIN AND A IS THE CO-DOMAIN

(FOR FUNCTION B TO A: Definition:

EACH ELEMENT OF B SHOULD HAVE UNIQUE IMAGE IN A, more than one element of B can be mapped to one single element of A but, one element of B should not be mapped to more than one element of A)

FOR THAT, 5 has 4 choices, 6 has also 4 choices

so total mapping=4*4=16

Mam I have a doubt If we have A={1,2,3,4,7} and B={5,6} then what is the total number of functions we get from A->B

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