Phillip D. answered • 11/19/19
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Given a list of ten books, from which he must choose two. This is the start of a combination / permutation problem. We must now determine whether this is a combination (order doesn’t matter (AB is the same as BA) or a permutation (order matters (AB is different than BA)). In this case order doesn’t matter because reading book A than book B is the same as reading book B than book A. We now use the formula for combinations: 10C2 = 10! / ((10-2)! * 2!) = 10! / (8! * 2!) = (10*9) / 2 = 45 ways.
Phillip D.
A less structured but intuitive way of looking at this problem is we have two book spots to fill. In the first spot, how many ways can we fill it? We can fill it 10 ways. Looking at the second spot, how many ways can we now fill it if we already used one before it? 10 - 1 = 9 ways. Using the fundamental principle of combinatorics / probability, we multiply both spot values to get 90 possible ways to arrangements. However we must remember reading book one then reading book two is the same as reading book two then book one (AB = BA), so we must get rid of the extra cases, which we do so by diving by two to arrive at 45 possible ways.11/19/19