in Set Theory & Algebra
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34 votes
34 votes

Suppose $X$ and $Y$ are sets and $|X| \text{ and } |Y|$ are their respective cardinality. It is given that there are exactly $97$ functions from $X$ to $Y$. From this one can conclude that

  1. $|X| =1, |Y| =97$
  2. $|X| =97, |Y| =1$
  3. $|X| =97, |Y| =97$
  4. None of the above
in Set Theory & Algebra
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  1. 97 functions.
  2. 1 function.
  3. 97^97 functions.
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5 Answers

43 votes
43 votes
Best answer
We can say $|Y|^{|X|} = 97.$ Only option A satisfies this. Still, this can be concluded only because $97$ is a prime number and hence no other power gives $97.$
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4 Comments

@Mohitdas , yes we can't have one many function, but here we have 97 “different functions", each function corresponding to one element of X and one of 97 elements.

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Sir function is a relation for each input it has one output

Here 1 input has 97 different outputs

And I'm not getting “different functions”

Y and X are 2 sets

Can you please draw so that I can visualize..
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Here

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14 votes
14 votes

I think this question is not right and I have not got the right explanation of this question anywhere.

Everywhere is concluding $|Y|^{|X|} = 97^1 $ only satisfying but if it is satisfying that

Number of functions means how many mappings possible from domain to codomain

So from above 5 can be mapped by 97 ways so,

option (A)

edited by

4 Comments

hii @pritishc please read my answer once as well as selected question's discussion once...

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@Rishi yadav you should read it again. There's not even any ambiguity here. Probably you're not getting the definition of function properly. Please refer a standard book and it should be clear.

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My bad sir i got my mistake
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5 votes
5 votes

How to know whether 97 is a prime or not ? 

Take sqrt of 97 = 10 (approx)

now check whether 97 is divisible by all prime numbers which are <= 10 

If no then it is a prime number ... 97 is not divisible by 2,3,5,7 .. so 97 is a prime number...

So 97 cannot be written as a product of 2 numbers(apart from 97*1)...And so 97 cannot be represented as nm , (apart from 971 ) ...

So only way we could have got 97 functions is using 971 which is possible only if 1 element is in domain and 97 elements in co-domain.

B) says number of fuinctions is 1.

C) says number of functions is 9797

1 comment

Please explain that how option a is even a function?

1 element in X connected to all (97) elements in Y ....i think it does not satisfy the function definition..

In this case shouldn't be the answer d
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3 votes
3 votes

 functions = $co-domaindomain^{domain}$ and number of function given 97 so this is possible only if co-domain = 97 and domain = 1

 

2 Comments

we can't map one to many relationship in function . here we have 97 choices to select 1 function from "domain to co-domain"
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Thanks
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Answer:

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