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I tried to solve "birthdays paradox" using the "direct way" but not the one mentioned here in one of the topics - tried my own, and as one of the solutions I figured out that there was a need to consider cases where not only couples share same birthday but maybe groups of three four five etc.

Well, there is a well known formula to count unique handshakes (unique couples) amount when we have as given n (people amount):

n*(n-1)/2

My question: Is there more general formula that I can use if I need to know how many unique groups of three, or unique groups of four, or five etc...are there for a given n Thank you

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    $\begingroup$ yeah you can use the concept of subsets of n ,${ n \choose k}$ $\endgroup$
    – user998997
    Feb 13, 2022 at 21:41

1 Answer 1

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Yes, the number of ways to choose $k$ people from a set of $n$ people is ${n\choose k}=\frac{n!}{k!(n-k)!}.$

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  • $\begingroup$ Thank you very much $\endgroup$
    – Igor
    Feb 13, 2022 at 21:53
  • $\begingroup$ @Igor ofc, no prob $\endgroup$ Feb 13, 2022 at 21:54

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