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

[@mukulgupta4008]

Suppose sequence is 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

[@Cheenta]

Very nice!

[@vanamshobharani5003]

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

[@RameshKnowledgeIndex]

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 1,3,4......n. 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 slot3,4,5,.....n.
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

[@Abhisruta]

Exactly.

[@RameshKnowledgeIndex]

Thank you brother

[@Cheenta]

Nice work!

[@yashvardhan6521]

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

[@nextgodlevel]

sir next video link?

[@RameshKnowledgeIndex]

Thanks lot sir

[@Cheenta]

Thank you for watching