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For testing purposes

Hello, I wonder if someone could post the solutions for the hw0

for Part 19, why is the binary string for that representation 00110110? I understand 00 becaues there are 2 elements in the first bin, and then 11 because there are 2 empty slots, and then 01 because its one element in the third bin, but then the last 2 binary digits 10 are ambiguous because "10" is being interpreted in a different way than "11" was being interpreted earlier..

Nevermind, I think I got it. It might be easier to think of this as 6 bins and 2 balls, as it was suggested in the note later. You can have first ball land in 5 and second in 6, or you can have first ball to land in 6 and second in 5 - and this are two separate cases. However, if first ball lands into 6, and second ball goes into 6 - only one case, assuming that we are throwing 1st ball first and then 2nd all the time.

And actually same goes for event B. Why are we assuming that (6, 6) should be counted once?

For part 4 in the end, I don't understand why |A| = 6. I counted that there are 8 ways where order matters, A={(4*,6), (5*,6), (6*, 6), (5*, 5), (5, 5*), (6, 6*), (4, 6*), (5, 6*)} (* means that one of the dice is marked or has a scratch). I also recall that professor said it in class that order does matter and (6*, 6) != (6, *6). Please let me know if this makes sense, thank you a lot!

I also don't understand part 19. How do you reason that the sum of these subsets would add up to (n choose k+1) or n*(n-1 choose k)?

Can someone help explain part 20 and why we start picking the lowest numbered element and how that helps us solve this proof? Thank you!

page revision: 0, last edited: 26 Sep 2013 17:59