Im having problems in this question, i have asked my friends and teachers but they dk either, please help!!
Let A = {1,2,3,4}. The number of functions f: S ---> S. Such that f(f(i) = i for all 1<= i <= 4 is?
Im getting the answer 24 (4!) by applying the condition that the function must be bijective as its inverse exists. But the answer is 10
No clue how you mean you got 24.
There are 10 solutions:
The first is just f(i) = i (trivial).
Then there are 6 solutions that only flip 2 numbers such as
f(1) = 2, f(2) = 1, f(3) = 3, f(4) = 4
Finally there are 3 solutions that flip 2 pairs of numbers such as
f(1) = 2, f(2) = 1, f(3) = 4, f(4) = 3
"Flipping" numbers is the only way to have f(f(i)) = i so there are no other solutions.
bonkarfanboy1 [#4]No clue how you mean you got 24.
There are 10 solutions:
The first is just f(i) = i (trivial).Then there are 6 solutions that only flip 2 numbers such as
f(1) = 2, f(2) = 1, f(3) = 3, f(4) = 4Finally there are 3 solutions that flip 2 pairs of numbers such as
f(1) = 2, f(2) = 1, f(3) = 4, f(4) = 3"Flipping" numbers is the only way to have f(f(i)) = i so there are no other solutions.
explain in valorant terms? here is what i did
https://imgur.com/abluRhm
also who tf uses i as a variable 😭
keep [#5]explain in valorant terms? here is what i did
https://imgur.com/abluRhmalso who tf uses i as a variable 😭
Your solution is wrong although subtly. You do not account for my 3 "double flip" functions but you have accounted for the trivial solution 3 extra times. A function acts on every element in the set (at least here it has to for f(f(i))=i to hold).
Here are my 10 solutions: https://imgur.com/a/WleLcVC
bonkarfanboy1 [#4]No clue how you mean you got 24.
There are 10 solutions:
The first is just f(i) = i (trivial).Then there are 6 solutions that only flip 2 numbers such as
f(1) = 2, f(2) = 1, f(3) = 3, f(4) = 4Finally there are 3 solutions that flip 2 pairs of numbers such as
f(1) = 2, f(2) = 1, f(3) = 4, f(4) = 3"Flipping" numbers is the only way to have f(f(i)) = i so there are no other solutions.
I got 24 because if f(i) is self invertible
it should be bijective(one one and onto) Before this question i didnt know applying this condition leads to overcounting, thanks for you help.
keep [#5]explain in valorant terms? here is what i did
https://imgur.com/abluRhmalso who tf uses i as a variable 😭
What kind of function is this bro? What exactly are you counting? You cant count the images one by one and count them as functions. Its lucky you got the answer
PP12123213123 [#6]tell me why u put 4! in the answer
(i) f(a) = a, f(b) = b, f(c) = c, f(d) = d ; 1 case
(ii) f(a) = b, f(b) = a, f(c) = c, f(d) = d ; 6 case
(iii) f(a) = b, f(b) = a, f(c) = d, f(d) = c ; 3 case1+6+3 = 10 case
I put 4! because every self invertible function should be bijective( one one and onto) so i applied its condition and got 4!. I didnt realise it would lead to overcounting
Hades_Loves_Rb [#12]so bassicly then you and then from there you get and thats that! hope this helped
It definitely did, thank you puchan
bonkarfanboy1 [#7]Your solution is wrong although subtly. You do not account for my 3 "double flip" functions but you have accounted for the trivial solution 3 extra times. A function acts on every element in the set (at least here it has to for f(f(i))=i to hold).
Here are my 10 solutions: https://imgur.com/a/WleLcVC
basically 4c1, 4c2 and then 3c1
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