I thought that the simplest expression without using a calculator would be $20C2 +20C3+ 20C4 ... 20C20$. I then attempted to simplify to $21C3 + 21C5 + 21C7 + 21C9 + 21C11 + 21C13 + 21C15 + 21C17 + 21C19 + 20C20$ but I cannot seem to simply any further. The answer in the book that I am using is 220-21 but I'm not sure how they got this and how I should go about similar questions in the future. Thanks for any help
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You can use the fact that $\sum_{i=0}^n {n \choose i}=2^n$ – Ivan Kaznacheyeu May 05 '22 at 07:30
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Have a look at the collection of subsets of the set of students (how much are there?). Then discard the subsets with less than 2 students. – drhab May 05 '22 at 07:48
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You should use the complement, ie from total ways of selection, subtract ways where $20,19,$ students are rejected
Since each student can either be selected or rejected, there are a total of $2^{20}$ possible selections.
From this we subtract $\binom{20}{20} + \binom{20}{19} = 21$
true blue anil
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