Question

Use the principle of induction to prove that the number of derangements of n,dn, can be given by the recursion formula
d1 = 0, d2 = 1, dn = (n-1)(dn-1 + dn-2) (n >= 3).

Answer 1

first make n=k is true and after you need to prove that this is true for n=k+1 too

Answer 2

d3=(3-1)(d2 +d1) =2*(1+0) =2

Answer 3

1st take like this n=0,1

Answer 4

i mean d1=(0-1)(d1-1+d2-2)

Answer 5

here we can solve like this (-1)(d1-1+d1-2)
=-d1+1+d1-2
=-1
and also d2=1,so,d2=(1-1)(d2-1+d2-1)
=0
so the answer is (0,-1)

Answer 6

oh sorry mistake

Answer 7

u ca take opposite as n=k so
here we can solve also like this
i mean do the opposite then like n=K+1
=0-1
=-1
n=k+1
=-1+1
=0

Answer 8

whats ur answer?

Answer 9

,,down12" check it please newly because when n=1 so than d1=(1-1)(---) =0*(---) =0
-so because we know that one indifferent number time zero will be equal zero

yes ?