Copied from last weeks homework. I would love to find a time that is good for all parties that are interested.

I would like to meet on Tuesdays from 3-5 and work on this homework as a group. As of right now, meeting in the Union is the best meeting point. I'll be seated next to Starbucks and will bring my book. Please call or text me at 317-605-6720 with any questions or comments. Hopefully we can work through some of these problems together. Ryan Hossler

Is there anyway that we could move the meeting to a different time? I work from 2:30-5:30 monday through thursday... You think maybe we could do it sometime in the evening on some day instead?

I can do whenever is easy. I work nights, but can move my schedule around some if needed. If there are more then just you and I, then it's whenever it is easiest for us all. Maybe monday nights?

Monday nights can work ...I work til 7pm Tues Wed and Thurs, so if we did meet then I wouldn't be able to make it.

I would be interested, and anytime Monday works for me.

Monday nights would also work for me

Anyone wanna meet tomorrow around the time of 3:30? Or anytime from 330-615 or 9-11? I've got the right idea, but am missing a few key points on this one. Just name the time and place while also bringing your book with you and i'm there.

Hey does anyone know if he drops any homeworks or not??

HW5MA375S10

6.1 - 14, 19, 28, 30, 32, 36, 38

Section 6.1

14. I'm gettting 4*13*12*11*10*9/(52,5) which is a number that is much too small. Help? Same problem for #19 if anyone wants to explain.

You have 13 kinds of denominations to choose from, and you need to pick 5 of them. Then for each of the 5 choices, you have 4 suits to pick the one card. That is the logic for the numerator, and you have the denominator correct.

But (13 choose 5)*4)/(52 choose 5) is very small (smaller then before) .0019 isn't logical.

Remember, you are choosing 4 suits 5 different times, so it should be 4^5

19.

28.

Do we need to account for C(80,7), the total number of ways to chose 7 out of the first 80 positive integers? If not, why?

-- Try to think of this problem like this: How many ways can the computer choose 11 numbers so that your 7 are in them? Now, how many ways can the computer choose 11 numbers? 330/1.03E13?

-- You can also use the total number of ways to choose seven of the first eighty positive integers in your solution. How many ways can you pick seven "correct" numbers?

30. Completely and utterly lost on how to do this.

-Is this problem different from number 28? How does the part "choosing five (but not six) numbers" make a difference?

32.

36.

38. for part b i'm not sure whether it's independent or not, does it matter if the first coin is heads or tails because either way you need 2 heads in a row