Hello, I have some questions on counting that I am struggling with, it would be nice if someone can explain. Thanks a lot! Please Register or Log in to view the hidden image! 1) How many different three-scoop cones can be made from vanilla, chocolate, and strawberry ice cream? 2) A grade 9 student may build a timetable by selecting one course for each period, with no duplication of courses. Period 1 must be science, geography, or physical education. Period 2 must be art, music, French, or business. Periods 3 and 4 must each be mathematics or English. How many different time tables could a student choose? I would say 3x4x2x1=24 different time tables, but the answer in my textbook is 18. Why 18 instead of 24? 3 A school has locker locks that have 40 positions on their dials and use a three-number combination. 3a) How many combination are possible if consecutive numbers can't be the same? Are there any assumptions that you have made? Number of possible combinations = 40 x 39 x ? For the 3rd stage, should I multiply by 39 or 38? And what assumptions have I made? Nothing, right? I just compute the possibilities... 3b) Assuming that the first number must be dialled clockwise from 0, how many different combinations are possible?
The proplem doesn't say whether you're allowed multiple scoops of the same flavour or not, and what is meant by "different." Is strawberry-vanilla-chocolate the same as chocolate-vanilla-strawberry or not? And when you take into account that the scoops can be piled in any orientation (one on top of another, two down and one on top, depending on the size of the cone itself) you get a helluva lot of possibilities. I would also say 24. My timetable also has around 4 periods a day on most days, but each period is 2 hours long Please Register or Log in to view the hidden image! For the third stage, multiply by 39, since the only requirement is that the third number is different from the second. Assumptions: The three dials use the same 40 numbers, the 40 numbers on each dial are different, the dials are working properly so you can access all the combinations, the author of the question didn't accidentally or deliberately give you false information to mislead you, you're tall enough to reach the dials, you have enough time to try all the possibilities, you're not afraid of someone asking you difficult questions about why you're trying all the possibilities... I don't see how this would change the problem, unless there's something about school locker combination locks I don't know about.
Yes, as you have written it, the answer is 24. If the 2nd period only had 3 choices, then the answer would be 18. I long ago learned that the answers in the book are not 100% reliable. Mistakes are made, misprints, or even the problem gets editted, but the answer doesn't. Confusing, I know, but what can you do?
Kingwinner... I have noticed you comming here a lot for help with your homework. I am perfectly willing to help you understand the problems, and the principles behind them, but I am not going to work them for you. Understood?
I hate wrong answers from textbooks! It's like promoting the wrong mathematical concepts and misleading students... Please Register or Log in to view the hidden image!
Here comes a really cheallenging one, I did it for extra practice but I failed to... 4) 10 people are to be seated at a rectangular table for dinner. Tanya will sit at the head of the table. Henry must not sit beside either Wilson or Nancy. In how many ways can the people be seated for dinner?
OK...Tanya's position is fixed, so she is effectively eliminated from your calculations. Now you need to figure out how many ways you can seat Henry, Wilson, and Nancy. I would start this part by looking at Henry first. After that, the last six can be seated at random.
Tricky! Since it's an extra practice question, I've done most of it for you, but left some key elements for you to fill in. I could also have got it wrong, so make sure you check my logic. Assuming a ten seat table (no empty seats!). Put Tanya at the head. Now, we have nine seats and nine guests. How many ways can the nine guests be seated at the nine seats? 9 factorial, right? Now the hard part. How many of those 362880 ways are illegal? I'd do this in parts. First part - Henry is between Tanya and either Wilson or Nancy (4 ways to do this). The other guests can be distributed in 7! ways. Second part - Henry is between Wilson and Nancy. How many ways to do this? The other guests can be distributed in 6! ways. Third part - Henry is between Wilson or Nancy and another guest (not Wilson, Nancy, or Tanya). How many ways to do this? The other guests can be distributed in 6! ways. The answer in the end (if all my logic and maths is correct - a big IF!) is 272160.
Hmm... Poincare's method is easier. I get a different answer using it, which makes me think there's a mistake in my previous post. But I can't find it!
It appears to be in one of your bold "How many ways to do this?", the method you outlined works fine. I think it's easier to use inclusion/exlusion if you want to start with 9! and remove the bad arrangements. You can also arrange the order the other 8 guests, then place henry (don't 'seat' the guests until you've decided where henry fits in). There will be two cases for the arrangement of the 8 with different numbers of placements of henry depending on whether william and nancy are together or not.
What do you mean by inclusion/exclusion? Is this the method I suggested, or something else? Interesting - at first glance, I thought that Nancy and/or Wilson in first or last place would also be a special case, but no. I get the same answer as the first way... still checking my way.
Got it. I had my "Third part" wrong by a factor of two (forgot to allow Henry's partners to switch places). The magic number is 42 x 7! = 211680.
Something else. If A is the event that Henry is beside Wilson and B is the event that Henry is beside Nancy then the total allowable arrangements is: (total arrangements) - (# of ways A can happen) - (# of ways B can happen) + (# of ways A and B can happen simultaneously). The novelty here is counting the # ways A can happen (likewise B) we don't care if B happens or not (likewise A), so it's a little simpler than trying to find the number of ways A can happen without B happening (and vice versa). There's some double counting, but it's removed in the 4th term above (or rather, "reinserted"). #of ways A can happen, treat Wilson and Henry as 1 person, arrange these "8" people around the table in 8! ways, times 2 for the order of henry and wilson. # of ways B can happen, symmetric same as above. # of ways A and B can happen, treat Wilson, henry and nancy as 1 person, arrange these "7" people around the table in 7! ways, times 2 for wilson-nancy on the left/right of henry. We get: 9!-8!*2-8!*2+7!*2, same as your answer. inclusion/exclusion for more events- http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html Yep, it's all in seating them after their clockwise order has been decided and Henry has been fit in.
4) I think I need to ask a question first! I don't understand the question itself! "Tanya will sit at the head of the table" What does it mean by head of the table? Are there 1 or 2 possibilities? Does that mean Tanya can sit at either of the 2 locations? (either one of the single seat sides, right? Since it's a rectangle, there are 2 single seat sides) And are people considered "next to" one another around a corner?
We've assumed that the head of the table is a fixed place (only one possibility). We've also assumed that "next to" applies around a corner.
