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Mathematics 23 Online
OpenStudy (kirbykirby):

Why is k choose (k+1) = 0?

OpenStudy (kirbykirby):

Why do we say \[\left(\begin{matrix}k \\ k+1\end{matrix}\right)=0\] Is it because in the notation C(n, r) we assume n >= r? How come we wouldn't say it's "undefined" or simply say it's 1/(k+1) I know this is may be a stupid question =\

OpenStudy (unklerhaukus):

there are no combinations

OpenStudy (kinggeorge):

Basically, suppose you have 4 things. Now how many ways can you choose 5 things from those 4? You don't have enough things to choose from, so it must be that 4C5=0. It's the same argument for any k, not just 4.

OpenStudy (anonymous):

But \(\binom n 0 =1 \) is defined.

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