Question

A very hard problem.. help plz?

Answer 1

So the randomly chosen function is not necessarily 1-1 from AUB to AUB, however, we're given that it is 1-1 from A to AUB and from B to AUB. So let's figure out the total number of ways that it can be 1-1 from A and B onto AUB. Then we can figure out that ways it can be 1-1 from AUB onto AUB.

Answer 2

1-1 means that no two inputs can yield the same output. Therefore, from A onto AUB, I have 5 inputs and 10 possible outputs. Considering the inputs 1 at a time, I have
10 choices for the first output.
9 choices for the second
8 choices for the third
7 choices for the fourth
6 choices for the fifth
This gives (10*9*8*7*6*5) ways for A to be mapped onto AUB.
To map B onto AUB, I have the same number of possibilities, so this gives a universe of:
(10*9*8*7*6)*(10*9*8*7*6)

Answer 3

Now IF AUB onto AUB is 1-1, then A onto AUB and B onto AUB are 1-1. So if we find the number of possible ways to make a 1-1 mapping from AUB to AUB, then those ways will all be within my universe of mappings that are 1-1 from A to AUB and from B to AUB.
The number of 1-1 mappings from AUB to AUB:
10 choices for input 1
9 choices for input 2
etc
Gives 10! mappings.

Answer 4

That gives me a probability of
(3628800)/(914457600)
Rewrite that in reduced form.

Answer 5

And then add the numerator and the denominator, for some arbitrary reason that the problem does not state. Lol.

Answer 6

yeah they love this add m and n business. not sure why

Answer 7

I actually feel like bringing in this whole "M and N relatively prime" business is just confusing the issue. These a probability and combinatorics problems. Stating them in that manner really feels like intentionally trying to confuse your students.