The number of 9-digit even natural numbers fo | vedclass

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Question

(C) $9$-digit number must start with $1$ (since it cannot start with $0$).Since the number is even,it must end with $0$.Let the number be $d_1 d_2 d_3

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) $9$-digit number must start with $1$ (since it cannot start with $0$).Since the number is even,it must end with $0$.Let the number be $d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9$.We have $d_1 = 1$ and $d_9 = 0$.Since no two consecutive digits are $0$,$d_8$ must be $1$.The number looks like $1 d_2 d_3 d_4 d_5 d_6 d_7 1 0$.For the remaining $6$ positions ($d_2$ to $d_7$),we need to fill them with $0$ or $1$ such that no two $0$s are consecutive.Let $a_n$ be the number of such sequences of length $n$. For $n=6$,we use the recurrence relation $a_n = a_{n-1} + a_{n-2}$.For $n=1$,sequences are $(0), (1) \rightarrow 2$.For $n=2$,sequences are $(0,1), (1,0), (1,1) \rightarrow 3$.For $n=3$,$a_3 = 3+2 = 5$.For $n=4$,$a_4 = 5+3 = 8$.For $n=5$,$a_5 = 8+5 = 13$.For $n=6$,$a_6 = 13+8 = 21$.Thus,there are $21$ such numbers.

The number of $9$-digit even natural numbers formed using only the digits $0$ and $1$,such that no two consecutive digits are $0$,is:

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