Forming 4-digit odd numbers under certain rules How many four-digit odd numbers can be formed such that every $"3"$ in the number is followed by a $"6"$?


A) 108
B) 2592
C) 2696
D) 2700


A "3" need not be immediately followed by a "6", and each "3" must be uniquely associated with a "6", i.e. multiple occurrences of a "3" cannot share the same "6".
Hint: firstly take a look at the possibilities without $"3"$: you have $8$ choices for the 1st digit, $9$ choices for the 2nd digit, $9$ choices for the 3rd digit, and finally $4$ choices for the last digit (in order to be odd, the number must end with $1,3,5,7 \text{ or } 9$) $\Rightarrow$ at least $8\cdot 9\cdot 9\cdot 4 = 2592$ numbers, so answers A and B can't be right...then try by examinate the number of numbers with only 1 $"3"$, 2 $"3"$, and so on...
consider ABCD , A{1..9} , B{0..9}, C{0..9} D{1,5,7,9}
P{odd}=9*10*10*4=3600
subtract P{A=3 and B not =6 }
subtract P{B=3 and C not = 6}
subtract P{c=3 }
add P{A=3,B=6,c not =3}
Answer=3600-1*9*10*4 -....... + ....=2592