Derangement and Recursion || Combinatorics || Math Olympiad || ISI Entrance

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In this video, we will learn the concept of derangement and recursion, a part of Combinatorics.
This is helpful for Math Olympiad and ISI Entrance.

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  1. Cheenta Academy for Olympiad & Research
  2. Derangement
  3. Recursion
  4. Combinatorics
  5. Math Olympiad
  6. ISI Entrance
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Suppose sequence is <a_n> for n>=1, Recursion forumla : a_n = a_{n-1} + n for n >=2 where a_1=1
and a_n = {n(n+1)}/2

Answer to the Question in Video

mukulgupta
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For triangular numbers we substitute the value of n and formula is [n(n+1)/2]

vanamshobharani
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A question of logic similar to that here of derangement came in iit jee 2014

yashvardhan
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In the second case 2 is not suppose to be on slot 1 so let us assume that slot 1 belongs to 2 [as slot 2 is fixed for 1]. Now you have n-1 numbers from 2 to n and n-1 slot Number of ways to dearrange them is Dn-1.
Thus the total number of derangement possible when position of 1 is fixed on slot 2 is Dn-1 + Dn-2.
We get the same result n-1 times if instead of slot 2 we fix 1 at
Therefore the total number of derangement possible is (n-1)(Dn-1 + Dn-2).
I felt to good after formulating such a rock solid argument

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