Let \(a_{n}\) 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 .

Question:
The value of \(b_{6}\) is

\(\because a_{n}=\) number of all \(n\) digit +ve integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 .
and \(b_{n}=\) number of such \(n\) digit integers ending with 1 \(c_{n}=\) number of such \(n\) digit integers ending with 0
Clearly, \(a_{n}=b_{n}+c_{n}\left(\because a_{n}\right.\) can end with 0 or 1)
Also \(b_{n}=a_{n-1}\) and \(c_{n}=a_{n-2}[\because\) if last digit is 0 , second last has to be 1]
\(\therefore\) We get \(a_{n}=a_{n-1}+a_{n-2}, n \geq 3\)
Also \(a_{1}=1, a_{2}=2\),
Now by this recurring formula, we get
\(a_{3} =a_{2}+a_{1}=3 \)
\( a_{4} =a_{3}+a_{2}=3+2=5 \)
\( a_{5} =a_{4}+a_{3}=5+3=8 \)
Also \(b_{6} =a_{5}=8\)