COMBINATORICS: Let A = {a1,a2,...,am} and B = {b1,b2,...,bn}. how many functions from A to B satisfy... assume that the functions are 1-1.
a)f(a1) = b1 and f(a2) = b2?
b)f(a1) = f(a2)?
c)f(a1) does not equal f(a2)?
d)f(a1) is an element of {b1,b2,b3}??

there are n!/(n-m)! 1-1 functions:
a) (n-2)!/(n-m-2)! ?
b) n choices for f(a1) and 1 option for f(a2) then n!/(n-m-2)! ?
c) n!/(n-m)!
d)3 choices for f(a1) therfore, 3n!/(n-m-1)! ?

Question

COMBINATORICS: Let A = {a1,a2,...,am} and B = {b1,b2,...,bn}. how many functions from A to B satisfy... assume that the functions are 1-1.
a)f(a1) = b1 and f(a2) = b2?
b)f(a1) = f(a2)? 
c)f(a1) does not equal f(a2)? 
d)f(a1) is an element of {b1,b2,b3}??

there are n!/(n-m)! 1-1 functions:
a) (n-2)!/(n-m-2)! ?
b) n choices for f(a1) and  1 option for f(a2) then n!/(n-m-2)! ?
c) n!/(n-m)!
d)3 choices for f(a1) therfore, 3n!/(n-m-1)! ?

anonymous · Answer

"...except that the functions are 1-1."

Is that supposed to say "assume" they're 1-1?

anonymous · Answer

yes

anonymous · Answer

For the second option, if $f:A\to B$ is supposed to be one-to-one, then $a_1=a_2$. Does this mean $a_2$ isn't mapped into $B$? The 1-1 assumption doesn't make sense to me here.

anonymous · Answer

But besides that, I'm pretty sure your other answers are correct.

anonymous · Answer

oh yea that doesnt make sense if its 1-1 f(a1) cant equal f(a2).

anonymous · Answer

thanks