Recursion or not

Probability Level 2

Can the famous derangement number be proved from recursion method? D ( n ) = n ! ( 1 2 ! − 1 3 ! + 1 4 ! − ⋯ + ( − 1 ) n n ! ) D(n)=\quad n!(\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 4! } - \cdots +\frac { { (-1) }^{ n } }{ n! } )

Yes None of these choices Can't be determined No

Prince Loomba by Prince Loomba , 21, India

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1 solution

Prince Loomba
May 10, 2016

For the proof by recurrence method click here

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