- Hey, can anyone tell me if I am doing number 6 right? I figured that it was just the product rule and the numbers being 4x6x6=144... Thoughts?
- In the problem regarding 5 consecutive letters, make sure you are counting each term only once. For example, a careless method would have AAAAAABCDE counted twice (once for the first set of 5 A and once for the second set of 5 A)
- In problem 40, how are you counting part a? I was doing 10 choose 5, but then I started thinking up other ways that seem like they could be right. Thoughts?
I did
$ 1*9*8*7*6*5 = 15120 $
because the bride has to be in one position so there is only one choice, and the rest of the positions can be any of the other people, but cannot repeat people (obviously) so decrease the number as you progress.
- Some problem with your assumption that the bride has only one choice, she actually has 6 choices. She can be in any of the 6 positions and the other people can be sit anywhere else. So your P(9,5) is correct but you need to multiply it by 6.
(I (someone else) can also verify this) Think of it as _A_B_C_D_E_ where she can sit in one of the 6 blanks, and the letters are filled by someone else.
- can someone tell me the meaning of "exactly one of .." such as in problem 40 and problem20
For example in one problem, it means you have the bride, or groom, but not both. So just use basic inclusion-exclusion_MA375Fall2008walther there.
If you find more than one way to count, and they both seem to be right, it's probably because they are both right. Just think of combinatorial proof_MA375Fall2008walthers.
