# Thread: If m=n find # of functions A onto B is m!. Using the string approach, would that be a

1. ## If m=n find # of functions A onto B is m!. Using the string approach, would that be a

If m=n find # of functions A onto B is m!. Using the string approach, would that be a satisfactory way of proving it to all sets with m=n? What do you think??

for ex)
input:__a(1) a(2)...___a(m)
choices:a(m) a(m-1)....a(1)= a(m)*a(m-1)*...*a(1)=m!

2. ## Re: If m=n find # of functions A onto B is m!. Using the string approach, would that

Originally Posted by Aquameatwad
If m=n find # of functions A onto B is m!. Using the string approach, would that be a satisfactory way of proving it to all sets with m=n? What do you think??
I am not sure what is meant by "a string approach".
But this is a simple permutation on the whole set, m! .

3. ## Re: If m=n find # of functions A onto B is m!. Using the string approach, would that

I don't know what the "string approach" is either, unless you mean "stringing together prior probability inverses to get the total probability" which seems to be what you're doing.

Also, please say finite set. This formula is not true in the general case (there are models of ZFC where although ${\aleph_0^\aleph^_0} = \beth_1$, $\beth_1 \ne \aleph_1$ (despite the fact it should be the correct answer in this case).