Consider the following code: Assuming no possibility of integer overflow, what will be returned by the following call to foo? foo(foo(3)+foo(4));
(15!)/(2!)
3!+4!
(7!)!
(3!+4!)!
15
This function computes factorial, except that it leaves out 2 from the series of multiplications.
So, foo(3)+foo(4) = 15.