1. Interpret the problem

We form $n$-digit positive integers using only digits $0$ and $1$, with the condition that no two consecutive digits are $0$.

Since the number is an $n$-digit positive integer, the first digit cannot be $0$. Hence every valid number must begin with $1$.

We define:

• $a_n$ = total number of valid $n$-digit numbers,
• $b_n$ = number of valid $n$-digit numbers ending in $1$,
• $c_n$ = number of valid $n$-digit numbers ending in $0$.

We need $b_6$.

2. Set up recurrences

A valid $n$-digit number ending in $1$ can be obtained by appending $1$ to any valid $(n-1)$-digit number. Therefore,

$b_n = a_{n-1} = b_{n-1}+c_{n-1}.$

A valid $n$-digit number ending in $0$ cannot have the previous digit as $0$, so it must come from a valid $(n-1)$-digit number ending in $1$. Hence,

$c_n = b_{n-1}.$

Thus,

$a_n = b_n + c_n.$

3. Find initial values

For $n=1$, the only valid 1-digit number is $1$. So,

$b_1=1, \qquad c_1=0, \qquad a_1=1.$

4. Compute successively up to $n=6$

For $n=2$

$b_2=a_1=1, \qquad c_2=b_1=1.$ So, $a_2=1+1=2.$

For $n=3$

$b_3=a_2=2, \qquad c_3=b_2=1.$ So, $a_3=2+1=3.$

For $n=4$

$b_4=a_3=3, \qquad c_4=b_3=2.$ So, $a_4=3+2=5.$

For $n=5$

$b_5=a_4=5, \qquad c_5=b_4=3.$ So, $a_5=5+3=8.$

For $n=6$

$b_6=a_5=8.$

5. Check with options

The value is

$b_6=8.$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.