Remember, Rule #5 says that if H works, then G must work on the following day. The contrapositive of this is that if G does not work on the day following H, then H does not work. Since we are told that G does not work, it would be impossible for G to work on the day following H. Therefore, H must not work.
This leaves F, I, and J to fill the 5 spots. What do we know about these variables? We know that the max that any one variable can appear is twice in the week, but not on consecutive days. This means that two of the variables will appear twice, and one will appear once. We also know from Rule #4 that F never works later in the week than J.
We could make J the only variable that appears once without breaking any of the rules as follows: F I F I J
We could make F the only variable that appears once without breaking any of the rules as follows: F I J I J
However, we cannot make I the only variable that appears once without breaking any rules. This is because we cannot separate the F from the other F and J from the other J while also making it so that both Fs appear before both Js. For example, if we tried F I F J J we would have an invalid scenario because the Js are consecutive, thereby breaking Rule #3. Therefore, it MUST be true that I appears twice, so (D) is the correct answer.
Does that make sense? Please let us know if you have any other questions and best of luck with your continued studies!
zacharylouiskane@gmail.comon June 29, 2020
For the fifth question to this game ... I started with the boxed rule number 5 [HG].... now we can have G anywhere in the diagram without the H NEEDING to be attached.... but we can't put H anywhere in the diagram if there is NO G suceeding correct? So four question 5 I will be placing 3 Fireman in 5 days ?
