$${6 \choose 2} - 1=14$$
I found this in a quantum computing paper and I cannot understand why the result is $14$. This looks like a vector, and I do not know how to properly treat a problem like this.
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This is unreadable. Link to paper? – Moo Apr 22 '21 at 12:40
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https://static-content.springer.com/esm/art%3A10.1038%2Fs41534-019-0240-1/MediaObjects/41534_2019_240_MOESM1_ESM.pdf – G.Chri Apr 22 '21 at 12:40
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if you can check the caption in figure S4 – G.Chri Apr 22 '21 at 12:41
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2I guess you mean ${6 \choose 2} - 1$ , right? – Matti P. Apr 22 '21 at 12:41
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yes, sorry for my post not being clear – G.Chri Apr 22 '21 at 12:42
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4$\binom{6}{2} = \frac{6\cdot 5}{2} = 15$ – Vishu Apr 22 '21 at 12:43
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3@G.Chri: You read that as a binomial 6-choose-2 = 15 - 1 = 14, see https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php – Moo Apr 22 '21 at 12:43
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Thanks a lot for your help! – G.Chri Apr 22 '21 at 12:45
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See this – Henno Brandsma Apr 22 '21 at 13:36
1 Answers
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The notation $\binom{n}{r}$ is an alternative (more common amongst people who've studied higher-level math) to the notation $^{n}\mathrm{C}_r$, representing the number of combinations of $r$ objects from $n$, i.e. the number of ways to choose $r$ different numbers from $\{1,\dots,n\}$ with order not being important.
The formula is $$\binom nr=\frac{n!}{(n-r)!r!},,$$ which for $n=6,r=2$ gives $15$.
Especially Lime
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