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If-n-objects-are-arranged-in-a-row-then-find-the-number-of-ways-of-selecting-three-of-these-objects-so-that-no-two-of-them-are-next-to-each-other-

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Question Number 21587 by Tinkutara last updated on 28/Sep/17
If n objects are arranged in a row, then find the number of ways of selecting three of these objects so that no two of them are next to each other.
$$\mathrm{If}\:{n}\:\mathrm{objects}\:\mathrm{are}\:\mathrm{arranged}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row},\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{selecting} \\ $$$$\mathrm{three}\:\mathrm{of}\:\mathrm{these}\:\mathrm{objects}\:\mathrm{so}\:\mathrm{that}\:\mathrm{no}\:\mathrm{two}\:\mathrm{of} \\ $$$$\mathrm{them}\:\mathrm{are}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$
Commented by mrW1 last updated on 29/Sep/17
Now I understand the question.
$$\mathrm{Now}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{question}. \\ $$
Answered by mrW1 last updated on 06/Oct/17
to select 3 objects from n objects there are totally C_3 ^n =((n(n−1)(n−2))/6) ways for 3 objects next to each other there are (n−2) ways for 2 objects next to each other there are 2(n−3)+(n−3)(n−4)=(n−2)(n−3) ways ⇒number of ways without two objects next to each other is therefore ((n(n−1)(n−2))/6)−(n−2)−(n−2)(n−3) =((n(n−1)(n−2))/6)−(n−2)^2 =(((n−2)(n^2 −7n+12))/6) =(((n−2)(n−3)(n−4))/6) =C_3 ^(n−2)
$$\mathrm{to}\:\mathrm{select}\:\mathrm{3}\:\mathrm{objects}\:\mathrm{from}\:\mathrm{n}\:\mathrm{objects}\:\mathrm{there} \\ $$$$\mathrm{are}\:\mathrm{totally}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{n}} =\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{6}}\:\mathrm{ways} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{3}\:\mathrm{objects}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}\:\mathrm{there} \\ $$$$\mathrm{are}\:\left(\mathrm{n}−\mathrm{2}\right)\:\mathrm{ways} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{2}\:\mathrm{objects}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}\:\mathrm{there} \\ $$$$\mathrm{are}\:\mathrm{2}\left(\mathrm{n}−\mathrm{3}\right)+\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{4}\right)=\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)\:\mathrm{ways} \\ $$$$ \\ $$$$\Rightarrow\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{without}\:\mathrm{two}\:\mathrm{objects} \\ $$$$\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}\:\mathrm{is}\:\mathrm{therefore} \\ $$$$\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{6}}−\left(\mathrm{n}−\mathrm{2}\right)−\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right) \\ $$$$=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{6}}−\left(\mathrm{n}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}^{\mathrm{2}} −\mathrm{7n}+\mathrm{12}\right)}{\mathrm{6}} \\ $$$$=\frac{\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{4}\right)}{\mathrm{6}} \\ $$$$=\mathrm{C}_{\mathrm{3}} ^{\mathrm{n}−\mathrm{2}} \\ $$
Commented by Tinkutara last updated on 06/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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