The number of nine-digit even natural numbers | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) nine-digit number must start with $1$ (since it cannot start with $0$).
Since the number must be even, it must end with $0$.
Thus, the number is o

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) nine-digit number must start with $1$ (since it cannot start with $0$).
Since the number must be even, it must end with $0$.
Thus, the number is of the form $1 . . . . . . . 0$.
Let the sequence be $a_{1}, a_{2}, \dots, a_{9}$ where $a_{1} = 1$ and $a_{9} = 0$.
Since no two consecutive digits are $0$, the digit $a_{8}$ must be $1$.
So the number looks like $1 . . . . . . 1 0$.
We need to fill the $6$ positions between the first $1$ and the last $10$ with $0$s and $1$s such that no two $0$s are adjacent.
Let $x_{n}$ be the number of such sequences of length $n$ ending in $1$, and $y_{n}$ be the number of such sequences ending in $0$.
For a sequence of length $n$ ending in $1$, the previous digit can be $0$ or $1$. So $x_{n} = x_{n-1} + y_{n-1}$.
For a sequence of length $n$ ending in $0$, the previous digit must be $1$. So $y_{n} = x_{n-1}$.
This gives $x_{n} = x_{n-1} + x_{n-2}$, which is the Fibonacci sequence.
For $n=1$ (sequence $1$), $x_{1}=1, y_{1}=0$.
For $n=2$ (sequence $11, 10$), $x_{2}=1, y_{2}=1$.
For $n=3$ (sequence $111, 110, 101$), $x_{3}=2, y_{3}=1$.
For $n=4$ (sequence $1111, 1110, 1101, 1011, 1010$), $x_{4}=3, y_{4}=2$.
For $n=5$, $x_{5}=5, y_{5}=3$.
For $n=6$, $x_{6}=8, y_{6}=5$.
For $n=7$, $x_{7}=13, y_{7}=8$.
For $n=8$, $x_{8}=21, y_{8}=13$.
Since we have $6$ middle positions to fill starting after the first $1$ and ending before the last $1$, we look at the number of valid sequences of length $7$ starting with $1$ and ending with $1$, which is $x_{7} = 13$ (if we consider the fixed $1$ at start and $1$ at position $8$).
Actually, the number of valid sequences of length $n$ starting with $1$ and ending with $1$ is $F_{n}$ (Fibonacci). For $n=8$, the number of ways is $21$.

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

Similar Questions

The value of ${2^n} \{ 1 \cdot 3 \cdot 5 \cdots (2n - 3) \cdot (2n - 1) \}$ is

The number of words formed from the letters of the word $RAJASTHAN$ by taking all the letters at a time in which vowels are alternate is:

The value of $\sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module