```Question 474524
<br> A natural way to answer the question would be "There are many ways." Or even answer it with the question "Why should I be concerned about the exact number of ways to answer the survey?" Unfortunately, sometimes we have to be concerned with these things. Say, people who work in the national statistics office will frequently need to do something like that if they want to find some interesting results worth reporting, which will in turn have important consequence for the decisions made by the government. Being a little bit convinced that this may be in fact important can help in accepting the effort of actually finding out the answer to the question.
<br> The key for answering these types of questions (Combinatorial problem) is to find a SYSTEMATIC way to do the whole process by hand. Once you are able to organise the process of writing down all the combinations using some kind of systematic procedure with a clear logic as for what goes where, then it will turn out that you can count the number of combinations simply by doing the whole process in your mind instead of actually doing it on paper.<br>
Any systematic way of writing down all the combinations will do for this purpose. My approach would be something like this:<br>
I will start by imagining that I already have with me all these ranked lists (each possible ranking is there and they are all unique) and then I will ask myself: How can I sort them? Well, why not take only the ranked lists with one and the same item being ranked first (say, crime) and make this to be my first group. Do the same for all the other possibilities when it comes to rank 1. This way I will create 8 groups. Clearly, these 8 groups will have equal number of lists in them since each group must contain the number of lists that is equal to the number of different ways you can rank 7 remaining items. This is very cool, since to count them all, all I need is to count the lists in one group and then multiply that by 8. Now focus on only one of these 8 groups and ask again the same question: How can I sort the lists here? All the lists here have one and the same item being ranked first. What to do now? Well, we could apply the same logic but this time for the second ranked item. How many different second ranked items could be  within one such group (where always one and the same item is ranked 1st)? Since one item is fully utilised for the first rank position, then there are only 7 possible items that could be found on the second position. Therefore, this single group could be sorted into 7 groups, each of which has one and the same item on the second position. This kind of sorting could be then applied to each of the original 8 groups and we can see that each of the eight groups can be split into 7 sub-groups, each of which will have to contain the same number of lists. This is now super cool since all I need to do to count all the lists is to count the lists in one of these 7 sub-groups and then multiply that number by 7 (to get the total in one of the 8 groups) and then by 8 (to get the overall total), or more directly, multiply that by 56.<br>
After a bit of experience, this process of thinking goes much faster and you will quickly realise that you can continue along the same lines of thinking to further split sub-groups into sub-sub-groups, where the number of lists in one of these sub-sub-groups will be needed to be multiplied by 6 to get the number of lists within its original sub-group, and for which we already know that needs to be multiplied by 7 and then by 8 to get the number of all lists.<br>
So, we are at the point where we know that the number of all unique lists is equal to the number of lists in one sub-sub-group (where first three positions are always the same; hence only 5 items remaining to be ranked from position 4 to 8) times 6 (the number of sub-sub-groups in a sub-group) times 7 (the number of sub-groups in a group) times 8 (the number of groups). But this in itself is a pattern and surely it must be true that if we continue, the total number of all lists will be equal to 1 times 2 times 3 times 4 times 5 times 6 times 7 times 8, or as we would do it in algebra: 1*2*3*4*5*6*7*8, or simply 8! (read: eight factorial). If you do not have this function (factorial) on your calculator then you can try typing 8! into the WolframAlpha search engine (www.wolframalpha.com). 8! is equal to 40320.<br>
Are you interested to know how many lists were in the sub-sub-groups?
There should have been 40320/(8*7*6), or simply 1*2*3*4*5, or 5!, which is equal to 120. That was quite an achievement - counting 120 and then multiplying it by 6, then by 7, and then by 8, is way easier than counting all 40320. Yet, it is obvious that the whole sorting thing was a rather lousy trick since it would have been many times more difficult to sort the lists into these sub-sub-groups than counting all of them. On the other hand, by pretending to sort them the way we did, we managed to count them all very quickly indeed. No?<br>
Finally, do pay attention that we did simplify the problem immensely since we made an unstated assumption that every family answering the survey will in fact use all 8 numbers (from 1 to 8) to give a unique rank to each of the 8 items. If some of the families surveyed think that some of these items should be of equal concern then we do not have the right answer. Allowing for any rank to any item (ties included), the answer to our problem will have to be different. Still, solving that problem is in fact a bit easier than solving the one we solved already. Try it!```