Mathematics - Permutation & Combination Question with Solution | TestHub

$\because a_n=$ 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, $a_n=b_n+c_n(\because a_n$ 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 $a_n=a_{n-1}+a_{n-2},n\ge 3$ Also $a_1=1,a_2=2$, Now by this recurring formula, we get $\begin{array}{rl}{a}_3& =a_2+a_1=3\\ a_4& =a_3+a_2=3+2=5\\ a_5& =a_4+a_3=5+3=8\\ \text{ Also }{b}_6& =a_5=8\end{array}$