I have a set A={20,20,35,20,50,60}.I want choose any 4 elements from the set.How many ways can I choose 4 elements.

As far I know if the elements of the set are distinct the number of ways to choose 4 elements will be $6\choose 4 $.Since in this problem the element 20 repeat 3 times some combination will be over-calculated.How can I deal with this situation.

Can anyone explain how I solve the above problem with intuition?

  • $\begingroup$ Actually $A=\{20,35,50,60\}$. If you want to maintain the repetition then you should not choose from a set, but from a tuple $\langle20,20,35,20,50,60\rangle$. $\endgroup$
    – drhab
    Commented Sep 20, 2015 at 17:59
  • $\begingroup$ Since A has repeated elements, it is usually referred to as a multiset. $\endgroup$
    – user84413
    Commented Sep 20, 2015 at 18:10
  • $\begingroup$ Does the order play a role ? $\endgroup$ Commented Sep 20, 2015 at 18:12

1 Answer 1


If I were to address this particular problem, I'd look at it ad hoc. That is to say, I would observe that at least one of the $20$ elements must be chosen. So:

  • How many different combinations have one $20$?
  • How many different combinations have two $20$s?
  • How many different combinations have three $20$s?

The answers to each of these three questions can be answered using ordinary combinatorial techniques, and then the answers added up to yield a final result of $7$.

Some general approaches exist (generating functions come to mind), but I don't offhand know of one that will work for large values without some computational aid or significant effort.


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