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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!

Zook
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user129048
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  • 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. – Zook May 06 '14 at 18:16
  • @user129048 Here is a similar question http://math.stackexchange.com/questions/773207/how-many-subsets-of-size-n1-can-we-have-so-no-two-of-them-have-intersection-o Perhaps you will find it helpfull. – gebruiker May 06 '14 at 18:21

2 Answers2

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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 $a\lt b$.

André Nicolas
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Hint: ${10\choose 3}$-(# "bad" choices).

vadim123
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