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
$ 1x9x8x7x6x5 = 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.
