Matematické Fórum

stuart clark

There are $20$ points in a circle $A_{1},A_{2},...A_{20}$ .Find the number of ways of selecting $4$ points such

that there are at-least two points in between any two selected points

Brano · 25. 02. 2013 21:05

Let us count those selections where we choose $A_1$ (as a pivot) and other three points. Each point forbids two spaces to the right so we are about to choose $3$ out of
$20\ \underbrace{-4\times 2}_{\text{forbidden}}\ \underbrace{-1}_{\text{pivot}}=11$
so we have
$\binom{11}{3}$
choices.
Now we can rotate this, so that the pivot will be $A_2,A_3,...$ . Each original selection corresponds to 4 pivoted selections so the answer is
$\binom{11}{3}\times 20\times\frac{1}{4}=5\binom{11}{3}=825$ .

stuart clark · 03. 03. 2013 15:18

↑ Brano:Thanks

Matematické Fórum

#1 03. 02. 2013 14:45 — Editoval stuart clark (03. 02. 2013 14:45)

permutations

#2 25. 02. 2013 21:05

Re: permutations

#3 03. 03. 2013 15:18

Re: permutations

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