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I've been trying to solve this problem, but i don't know what technique i should be applying and my answers are coming out to be wrong. I've tried putting them in different groups but non of them give me the answer.

Question: There are 10 people standing in a straight line. Find the number of ways to choose 3 of them such that the no 2 of them consecutive.

Please help!

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marked as duplicate by David K, Mike Pierce, graydad, Micah, Jonas Meyer Jul 3 '15 at 2:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You can count the solutions if you reduce 10 to a smaller number. Look for a pattern. It's combinatoric, so look in Pascal's Triangle. $\endgroup$ – Zook May 6 '14 at 18:16
  • $\begingroup$ @user129048 Here is a similar question math.stackexchange.com/questions/773207/… Perhaps you will find it helpfull. $\endgroup$ – gebruiker May 6 '14 at 18:21
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A neat way to do the counting is to imagine $7$ $\ast$, the unchosen people, lined up like this: $$\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast$$ These determine $8$ gaps ($6$ of them between consecutive $\ast$, plus the $2$ "endgaps") where the chosen people might have been. The number of ways to choose $3$ people, no two adjacent, is the number of ways to choose $3$ of these gaps. This number is $\binom{8}{3}$.

Remark: More generally, the number of ways to choose $k$ people from a lineup of $n$, with no two adjacent, is equal to $\binom{n-k+1}{k}$, with the usual convention that $\binom{a}{b}=0$ if $a\lt b$.

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  • $\begingroup$ Thank a lot! thats a really neat way to solve this problem and I'll be sure to remember the formula too! $\endgroup$ – user129048 May 6 '14 at 18:52
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Hint: ${10\choose 3}$-(# "bad" choices).

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  • $\begingroup$ How do i find all the bad choices? $\endgroup$ – user129048 May 6 '14 at 18:35

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