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Let say we have a finite set, A, with the cardinality $n$, and want to find how many different subsets with the cardinality $r$ does A have.

By JUST using the product rule, which says that;

If I have a procedure consisting of $k$ tasks and each task, $t_i$, can be done in $n_i$ different ways, then the whole procedure can be done in $\prod_{i=1}^{k} n_i$ ways.

So, the task is choosing an element from the set A rah times,so for the first choice there is $n$ option, for the second choice there is $n-1$ option .., and for the last option we have $(n-r+1)$, so by the product rule there is $\frac{n!}{(n-r)!}$, but this is a permutation, not combination, so where is my mistake ?

Moreover, in order to find the total number of subsets of A, by the sum rule, I computed $\sum_{r=0}^{k} \frac{n!}{(n-r)!}$, whose result is given by Mathematica as a function of Gamma function.

Note that I checked the question, which was my main question, but the example that I gave is enough for resolving the confusion.

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  • As you said yourself, with your reasoning you are actually counting permutations. That is, each combination of $r$ items is counted $r!$ times, once for each possible permutation of the $r$ items. So to get the number of combinations you need to divide by $r!$. – Evangelos Bampas Feb 20 '17 at 13:11
  • @EvangelosBampas I don't understand how I'm counting permutation, I'm just directly applying the product rule – Our Feb 20 '17 at 13:14
  • Let's say you have a set with 2 elements, X and Y, and you're asking how many subsets of cardinality 2 there are. (Note: we already know the answer to this problem: there is just 1 subset with 2 elements). Let's apply the product rule: we have 2 choices for the first element (i.e., either X or Y) and 1 choice for the second (the remaining element). So what we're actually counting is the two following sequences of choices: (X,Y) and (Y,X). But we just wanted to count the unique 2-element subset {X, Y}. We counted {X, Y} twice, once for the permutation (X, Y) and once for the permutation (Y, X). – Evangelos Bampas Feb 20 '17 at 13:28
  • @EvangelosBampas ok, I get it, but the result it obviously wrong, but there is nothing in the definition of the product rule that doesn't allow me to use it in such a problem.I mean result is wrong, but WHY ? – Our Feb 20 '17 at 14:57
  • Of course you can use it. The question is what result it gives you. What I'm trying to say is that, just using the product rule, you end up counting the number of ordered lists of $r$ elements out of $n$. But if you just want the number of subsets of size $r$, you don't care about ordering. I'm not sure that I can give a better explanation, maybe someone else could take a stab at it. Also check out the answers to this question: How to know if its permutation or combination? – Evangelos Bampas Feb 20 '17 at 15:09
  • @EvangelosBampas No, actually I get, but I find it odd that none of the sources states this fact. Of course, you might counter it by saying that did I even ask it :) – Our Feb 20 '17 at 15:20

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