I did the question by making cases where f(x)=x or {f(x)=y,f(y)=x} or where connecting x to f(x) forms a closed cycle of 4. Does the given solution consider this? Or am I committing a flaw here?
#Number of functions question
sturdy dew
sturdy dew
@pale phoenix Can you ping @Apu?
pale phoenix
<@&1227988399579730072>
brazen snow
I did the question by making cases where f(x)=x or {f(x)=y,f(y)=x} or where conn...
what website is this
solar ruin
What is this insane question? 
sturdy dew
This was the case-working I did...
Well, the answer might be wrong, 442 is too little of a value for something like this.
thin wigeon
This was the case-working I did...
What is the use of f(x)=f'(x)
Why have they given it
thin wigeon
This was the case-working I did...
Why the 3! In the 4 1 1 1 and 4 2 1 cases
sturdy dew
Why the 3! In the 4 1 1 1 and 4 2 1 cases
Because the "cycle" can have a circular permutation in 3! ways.
thin wigeon
Because the "cycle" can have a circular permutation in 3! ways.
I still dont understand
sturdy dew
I still dont understand
Like for instace if f(1)=2, f(2)=3, f(3)=4 and f(4)=1, then you can write it out as 1->2->3->4->1 in a cirular way right? Then each cyclic permutation of this (for instance 1->3->2->4->1) gives you all other possible "cycles" of choosing function for some 4 numbers.
thin wigeon
Like for instace if f(1)=2, f(2)=3, f(3)=4 and f(4)=1, then you can write it out...
uh makes sense
yeah your answer seems right they didnt even consider 4 2 1 case
sturdy dew
+solved @thin wigeon