magine you have a passcode lock with 5 dials: ABCDE, on first dial you have 0-5, second 0-4… and last just 0 and 1. Random variable requires permutations that add up to 5 and how many cases are there.
For 6 numbers including 0 with at most 5 spots in an adding equation that’s relevant to our problem, there are the following ways for them to add up to be 5:
5+0, 4+1, 3+2, 3+1+1, 2+3, 2+2+1, 2+1+1+1, 1+1+1+1+1
So for each of the 8 case there are several events that satisfies the problem requirement:
Case 1: 5A - 1 event
Case2: 4As or 4Bs, both of them have rest of the 4 letters that can slide into the 5 spaces in front, in the middle or behind - 40 events i.e. AAACA or BEBBB
Case8: p5,5 =120 events
So here we arrived at the conclusion that Answer A or B is plainly wrong.
Answer C tells you to itemize each A as A1 - A5 and B and so on, and choose 5 from them to form a set. But when you have A1, B, C, D, E as a combination, how is it differ from A2, B, C, D, E as a combination?
Answer D tells you to first itemize each each A as A1 - A5 and B and so on, and choose 5 and arrange every 5 in some order. But when you have A1A2CDE, how is it differ from A2A1CDE?
So I would choose answer E.
(and in my analysis obviously case 2+3 and 3+2 has some overlaps that requires you to get rid of, so it is quite a complicated problem if it’s not a multiple choice problem.)
For 6 numbers including 0 with at most 5 spots in an adding equation that’s relevant to our problem, there are the following ways for them to add up to be 5:
5+0, 4+1, 3+2, 3+1+1, 2+3, 2+2+1, 2+1+1+1, 1+1+1+1+1
So for each of the 8 case there are several events that satisfies the problem requirement:
Case 1: 5A - 1 event
Case2: 4As or 4Bs, both of them have rest of the 4 letters that can slide into the 5 spaces in front, in the middle or behind - 40 events i.e. AAACA or BEBBB
Case8: p5,5 =120 events
So here we arrived at the conclusion that Answer A or B is plainly wrong.
Answer C tells you to itemize each A as A1 - A5 and B and so on, and choose 5 from them to form a set. But when you have A1, B, C, D, E as a combination, how is it differ from A2, B, C, D, E as a combination?
Answer D tells you to first itemize each each A as A1 - A5 and B and so on, and choose 5 and arrange every 5 in some order. But when you have A1A2CDE, how is it differ from A2A1CDE?
So I would choose answer E.
(and in my analysis obviously case 2+3 and 3+2 has some overlaps that requires you to get rid of, so it is quite a complicated problem if it’s not a multiple choice problem.)