Q. 1.8 - A First Course in Probability

Question

Consider $n$-digit numbers where each digit is one of the $10$ integers $0, 1, . . . , 9$. How many such numbers are there for which

(a) no two consecutive digits are equal?

(b) $0$ appears as a digit a total of $i$ times, $i = 0, . . . , n$?

Step-by-Step Solution

Answer

(a) Total numbers for which no two consecutive digits are equal are $10\times 9^{n-1}$.

(b) Total numbers for which $0$ appears as a digit a total of $i$ times, $i = 0, . . . , n$ are $\left(\begin{array}{c}n\\ i\end{array}\right)9^{n-i}$.

Part (a) Step 1. Given information.

We have to form $n$-digit numbers where each digit is one of the $10$ integers $0, 1, . . . , 9$ and no two consecutive digits are equal.

Part (a) Step 2. Find the numbers for which no two consecutive digits are equal.

Total no. of digits $=10$

So, the first place can be filled in $=10! \text{ways}$ $=10$

Each of the remaining $n-1$ place can be filled in $=9!\text{ways}=9$ as two consecutive digits cannot be equal.

Therefore, the total numbers for which no two consecutive digits are equal are $10\times 9^{n-1}$.

Part (b) Step 1. Given information.

We have to form $n$-digit numbers where each digit is one of the $10$ integers $0, 1, . . . , 9$ and $0$ appears as a digit a total of $i$ times, where $i = 0, . . . , n$.

Part (b) Step 2. Find the numbers for which 0 appears as a digit a total of i times.

No. of choices of $i$ places to put $0$ is $=\left(\begin{array}{c}n\\ i\end{array}\right)$

The remaining $n-i$ position can be filled by any of the $9$ digits, $1,2,........,9$.

Therefore, the total numbers for which $0$ appears as a digit a total of $i$ times, $i = 0, . . . , n$ $=\left(\begin{array}{c}n\\ i\end{array}\right)9^{n-i}$.