by Guest » Sat Jun 01, 2019 7:20 am
No, the correct answer is not C. The correct answer is E, 120.
To do this problem you will, of course, need to know what "bijection" means! A function is a "bijection" if it is both a "surjection" (onto) and an "injection" (one to one). The first means it "covers" all members of the range space. The second means that two different members of the domain space are not mapped to the same member of the range space.
Here, both domain and range are X= {1, 2, 3, 4, 5}. Start with "1". What can "1" be mapped into? The answer is, of course, any of "1", "2", "3", "4", or "5". That gives us 5 different functions right there. Once that is chosen, what can "2" be mapped to? Any of the 4 remaining members that "1" was not mapped to! Then, of course, there are 3 members remaining that "3" can be mapped to, 2 members remaining that "4" can be mapped to, and 1 member remaining that "5" is mapped to. That is a total of 5(4)(3)(2)(1)= 5!= 120 different choice so 120 different functions.