Let n denote the number of all n-digit positive integers formed by the digits `0, 1` or both such that no consecutive digits in them are 0. Let `b_n` = the number of such n-digit integers ending with digit 1 and `c_n`= the number of such n-digit integers ending with digit 0. The value of `b_6`, is

Options

A. 7
B. 8
C. 9
D. 11

Explanation

To solve for \( b_6 \), which is the number of 6-digit positive integers formed by the digits 0 and 1, where no two consecutive digits are 0, we can break down the problem step by step.

### Step 1: Understanding the Definitions
- Let \( a_n \) be the total number of n-digit integers formed by the digits 0 and 1 without consecutive zeros.
- Let \( b_n \) be the count of n-digit integers ending with 1.
- Let \( c_n \) be the count of n-digit integers ending with 0.

### Step 2: Establishing Relationships
From the definitions:
- Any n-digit number can either end with 1 or 0.
- If it ends with 1, the previous digit can be either 0 or 1.
- If it ends with 0, the previous digit must be 1 (to avoid consecutive zeros).

Thus, we can establish the following relationships:
- \( b_n = a_{n-1} \) (if the last digit is 1, the first \( n-1 \) digits can be any valid configuration)
- \( c_n = b_{n-1} \) (if the last digit is 0, the previous digit must be 1)

### Step 3: Total Count
The total number of n-digit integers can be expressed as:
\[ a_n = b_n + c_n \]

### Step 4: Recurrence Relations
Substituting the relationships into the total count gives:
\[ a_n = b_n + c_n = b_n + b_{n-1} \]
This means:
\[ a_n = b_n + b_{n-1} \]

### Step 5: Base Cases
Now, let's establish some base cases:
- For \( n = 1 \):
- The only valid 1-digit number is 1. Thus, \( a_1 = 1 \), \( b_1 = 1 \), \( c_1 = 0 \).

- For \( n = 2 \):
- The valid 2-digit numbers are 10 and 11. Thus, \( a_2 = 2 \), \( b_2 = 1 \), \( c_2 = 1 \).

- For \( n = 3 \):
- The valid 3-digit numbers are 101, 111, and 110. Thus, \( a_3 = 3 \), \( b_3 = 2 \), \( c_3 = 1 \).

### Step 6: Calculating Further Values
Using the recurrence relations, we can calculate further values:
- For \( n = 4 \):
- \( b_4 = a_3 = 3 \)
- \( c_4 = b_3 = 2 \)
- \( a_4 = b_4 + c_4 = 3 + 2 = 5 \)

- For \( n = 5 \):
- \( b_5 = a_4 = 5 \)
- \( c_5 = b_4 = 3 \)
- \( a_5 = b_5 + c_5 = 5 + 3 = 8 \)

- For \( n = 6 \):
- \( b_6 = a_5 = 8 \)
- \( c_6 = b_5 = 5 \)
- \( a_6 = b_6 + c_6 = 8 + 5 = 13 \)

### Conclusion
Thus, the value of \( b_6 \) is:
\[ \boxed{8} \]