Aman Bansal
Last Activity: 12 Years ago
Dear Rahul,
ts say A = {a,b,c,d} and B = {s,t,u}. we want to find the number of onto functions f:A→B.
well, some member of B has 2 pre-images. there are 3 ways (= n) this can happen. having made THAT choice, we have to count the number of ways we can pick two elements of A to go to that member of B. so we need to know how many ways there are to pick 2 things out of 4 things, that number is "4 choose 2" or:
4!/(2!(4-2)!) = 4!/(2!2!) = 24/((2)(2)) = 24/4 = 6 (this is the "stuff inside the parentheses", which is called a binomial coefficient).
having decided which 2 elements of A get "doubled up" onto which element of B, we now have to assign the remaining targets for the rest of A.
we have 2 choices for where to send the 3rd element of A (since one element of B is already spoken for), after which our last choice is already made for us.
so we have (2)(1) = 2! ways to assign the rest of A, for a grand total of 3(6)(2) = 36 possible onto functions.
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Aman Bansal
Askiitian Expert