Submitted by Heirio on Tue, 09/05/2017 - 20:29
Forums:
Okay, I've come across a method which may or may not be valid and doesn't involve the adding of fractions (I don't think it does, anyway).
Essentially, I label my two n's n1 and n2.
n1 can be in m + 1 positions, since it cannot be at the start but can be anywhere else. So n1 goes with m + 1
n2 cannot go right next to n1, so its range of positions is m - 1.
So I basically said that the total range of spaces for the n's is m + 1, making that the denominator. I said that since it can only work with m - 1 spaces, so this is the numerator.
Thus, I got (m - 1)/(m + 1).
My gut tells me that while it is the right answer, the logic behind it is weird.
Could you tell me if this is right or not?
I don't think this is a valid
I don't think this is a valid method to get to the answer. For example, your 2nd person with a £2 coin ('$n_2$') can be next to your 1st person ('$n_1$') so long as you already have greater than or equal to two £1 coins.
I see, so due to the nature
I see, so due to the nature of the preparation and the first part, I'm guessing the proper method involves adding fractions with different denominators? I'm also guessing I'd need to use the answer to part i?
Preparation
There is a (deliberate) connection with the preparation (q2 i b IIRC). You are told to consider the first three people in the queue, and that there are exactly 2 people with a £2 coin.
The options for the first 3 people are:
£1, £1, £1
£1, £1, £2
£1, £2, £1
£2, £1, £1
£1, £2, £2,
£2, £1, £2
£2, £2, £1
In some of these cases you will be able to sell tickets to everyone, and in others you cannot. Sum the relevant ones!