I was just making a joke. This round was pretty rough in a few places, I haven't taken statistics or linear algebra so a good chunk of that was beyond what I knew naturally, and I couldn't even find anything related to #13. I guess everyone else managed better than I did on that one since I had no clue but to guess.

Oh, I know you were joking, don't worry. :)

Just making sure then, no harm done. Go Team U.S. Residents Not Getting Pissed At Each Other For A Change!

Points will be awarded at the end of the round, More easier for me to keep track.

LOL. True enough. :)
OK. Thanks.

3-
The point at -3 is full, so it includes -3 and then grows to the positves
x>= -3

8-
9 letters in total, 4 repetions of e, 2 repetions of n
9!/(4!2!) = 7560

9-I had this wrong, and solved like 8

12!/10! = 12*11 = 132

Now I see, I think I didn't discount the repetions of the extra questions

13-
nCk = nC(n-k)
100C2 = 100C98

(For those that don't know, in selecting k items from a group of n, the difference between permutations [P(n,k)] and combinations [C(n,k)] is that in permutations, order matters. In combinations, it does not.)

P(n, k) = n! / (n-k)!
C(n, k) = n! / (k! * (n-k)!)

3. Based on the graphs in other problems, we assume the blue colored region to the right of -3 is the area of interest. A closed dot indicates -3 is part of the solution. Therefore, (C) x ≥ -3

8. There are 9 letters in the word seventeen, however, only 5 of them are unique; there are 4 e's and 2 n's. This means that while there are 9! ways to arrange the letters, a significant number of those rearrangements are redundant. In every rearrangement, the e's and n's can be shifted around, in 4! ways and 2! ways, respectively. Since this occurs in every arrangement, they are divided out. 9! / (4! * 2!) = (B) 7,560

9. The question is unclear as to whether order matters on this test. However, given that P(12, 10) = 12! / (12-10)! = 239,500,800 does not appear among the answer, we must assume it does not. Therefore, the answer is C(12, 10) = 12! / (10! * (12-10)!) = 12! / (10! * 2!) = (B) 66

13. C(100, 2) and C(100, 98) both describe the exact same problem. Think about it in terms of splitting a group of 100 items into 2 groups: If you choose 2 items out of 100, then you are by definition leaving behind a second group, 98 out of 100, and vice versa. Therefore, if C(100, 2) is defined as 4,950, then C(100, 98) is (A) 4,950

zotofmu's post pretty much is the same process I followed.

So i decided to give you tickets like this
-------------------------------------------------------------------------------------------------------------- ----------------------
If you got all answers right you get a total of 400 tickets
if you got some questions wrong, Subtract 5 tickets for each wrong question,
-------------------------------------------------------------------------------------------------------------- ----------------------
Now here's the fun part,
I am doing this the honor code way,
You tell me what you earned.
And well if you are being honest, Who knows you may get a few extra tickets,
Or if your being dishonest, Maybe a few less tickets
-------------------------------------------------------------------------------------------------------------- --------------------------------------------

Anyway
-------------------------------------------------------------------------------------------------------------- ----------------------
Answers for Part 3
-------------------------------------------------------------------------------------------------------------- --------------------------------------------
1. D
2. D
3. B
4. A
5. B
6. C
7. D
8. D
9. A
10. A
11. A
12. C
13. B
14. D
15. C
16. A
17. D
18. B
19. A
20. C

-------------------------------------------------------------------------------------------------------------- --------------------------------------------
Answers for part 4
-------------------------------------------------------------------------------------------------------------- --------------------------------------------
1. B
2. D
3. C
4. C
5. D
6. A
7. C
8. B
9. B
10. C
11. B
12. A
13. A
14. B
15. D
16. D
17. B
18. B
19. C
20. D
-------------------------------------------------------------------------------------------------------------- ----------------------
Check Previous posts for Part 1 and 2
-------------------------------------------------------------------------------------------------------------- ----------------------

Part 1: 20 correct (+5 extra for checking the answers and tell you if we got any wrong)
Part 2: 20 correct (+5 extra "for bringing everything back on topic, +2 for right answers)
Part 3: 20 correct
Part 4: 20 correct
Makes 412 tickets...

Also, I thought this round was alright. It was a bit of a challenge, but nothing over-the-top extremely difficult once I remembered how to do things. (It has been awhile for me.) I have no suggestions for the next round.

Part 1: 19 correct
Part 2: 20 correct
Part 3: 20 correct
Part 4: 20 correct

So, I think 395 tickets?

This round wasn't bad, but I don't have any ideas for the next round(s).

Part 1: 19 correct (+5 extra for checking answer, if still valid)
Part 2: 19 correct
Part 3: 19 correct
Part 4: 19 correct

My suggestion: Please not do a contest in story writing or similar. I'm Italian, so i'm not able to write as an american/englishman in english :(

Multiple choice test, like this, are the best solution.