Say, the position of \(n\) is \(k^{th}\) from beginning.

\(k-1\) numbers would be in left of\(n\), and \(n - k\) numbers would be to its right, Select \(k-1\) numbers in the left in\({{n-1} \choose {k-1}}\) ways and arrange them in one order(increasing). Now the others would automatically be arranged in right of \(n\) in descending order in \(1\) way.

Hence , we conclude that :

No. of ways =\(\displaystyle \sum_{k = 2}^{n - 1} {{n-1} \choose {k-1}} = \boxed{2^{n-1} - 2}\)
–
Jatin Yadav
·
3 years, 9 months ago

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TopNewestSay, the position of \(n\) is \(k^{th}\) from beginning.

\(k-1\) numbers would be in left of\(n\), and \(n - k\) numbers would be to its right, Select \(k-1\) numbers in the left in\({{n-1} \choose {k-1}}\) ways and arrange them in one order(increasing). Now the others would automatically be arranged in right of \(n\) in descending order in \(1\) way.

Hence , we conclude that :

No. of ways =\(\displaystyle \sum_{k = 2}^{n - 1} {{n-1} \choose {k-1}} = \boxed{2^{n-1} - 2}\) – Jatin Yadav · 3 years, 9 months ago

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