How many four-digit odd numbers can be formed such that every $"3"$ in the number is followed by a $"6"$?
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".
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
On
total no. of digits without 3 is 2592. let us assume the number is "x x x y". where y is an odd digit but not 3, since every 3 has to be followed by a 6. hence y can be 1,5,7,9. also, 3 cannot fill into the third x because if it does so, then 6 cannot follow it. now there are three cases.
case 1:
3 6 x y
there are 4 possibilities for y and 9 for x. so it can be done in 9*4= 36 ways.
case 2:
x 3 6 y
here too, y can take 4 forms and x can take anything but 0 and 3, hence 8 forms.
there are 4*8= 32 possibilities
case 3:
3 x 6 y
y can take 4 forms and x can take 9 (0,1,2,4,5,6,7,8,9) forms
number of ways of filling it up is 4*9= 36 possibilities.
adding the number of numbers without 3 with those that have three, we get, 2592+36+36+32= 2696, which is the answer.
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...