# How many surjective functions are there when mapping a n-set to a k-set?

Is there some kind of relation, like maybe a recurrence relation or something?

### 4 Answers

- Anonymous7 years agoBest Answer
My favorite part of taking calculus has been forgetting it.

I still use algebra fairly regularly in my adult life, but I have never found a practical application for calculus in any job or hobby I've had.

- CAustinLv 67 years ago
It's going to be kCn; you're taking a set of size n and choosing positions for them out of a set of size k. (That's n!/(k!*(n-k)!), analytically.)

If you're trying to figure out the analytics of that combinatoric process for the first time, you can think of it as taking the elements from set n one by one, assigning them a mapping in set k, and then going to the next one until you run out of elements in set n. In this process, there are always k - i + 1 mapping possibilities for the element from set n where i is the iteration index of the process. Solve the resulting sigma expression and you come up with the normal kCn expression.

Source(s): Also, I agree with the people up top: you'll get more answers by posting this in Mathematics. I just happen to work with the stuff :) http://en.wikipedia.org/wiki/Binomial_coefficient If you scroll down to the "computing the value" section just under the contents, I think you're thinking of the recursive formula they describe there. The factorial formula, which I described, it just a quicker way of evaluating the same thing. - Anonymous7 years ago
n-set to the whatnow ?

Source(s): you know youre in trouble when not only do you not know the answer - but have no idea what the question is !