Q. 1.10 - A First Course in Probability

Question

How many $5$-digit numbers can be formed from the integers $1, 2, . . . , 9$ if no digit can appear more than twice? (For instance, $41434$ is not allowed.)

Step-by-Step Solution

Answer

$5$-digit numbers that can be formed are $52920$.

Step 1. Given information.

We have to form $5$ digit numbers using the integers $1, 2, . . . , 9$.

No digit can appear more than twice.

So, there will be three mutually exclusive even in which this can happen.

Step 2. Find the number of ways in which all 5 digits appear only once.

The no. of ways of choosing first digit is $\left(\begin{array}{c}9\\ 1\end{array}\right)=9$

The no. of ways of choosing first digit is $\left(\begin{array}{c}8\\ 1\end{array}\right)=8$

The no. of ways of choosing first digit is $\left(\begin{array}{c}7\\ 1\end{array}\right)=7$

The no. of ways of choosing first digit is $\left(\begin{array}{c}6\\ 1\end{array}\right)=6$

The no. of ways of choosing first digit is $\left(\begin{array}{c}5\\ 1\end{array}\right)=5$

Therefore, the number of ways in which all $5$ digits appear only once = $9\times 8\times 7\times 6\times 5=15120$

Step 3. Find the number of ways in which one digit appears twice.

No. of ways in which we can select the $1$ digit that will appear twice is $=\left(\begin{array}{c}9\\ 1\end{array}\right)=9$

No. of ways in which we can select 2 spots to place the 1 digit that will appear twice is $=\left(\begin{array}{c}5\\ 2\end{array}\right)=10$

The no. of ways of selecting $3$ digits out of the remaining $8$ digits is $=8\times 7\times 6=336$

Therefore, the number of ways in which one digit appears twice $=9\times 10\times 336=30240$

Step 4. Find the number of ways in which two digits appear twice.

No. of ways in which we can select the $2$ digits that will appear twice is $=\left(\begin{array}{c}9\\ 2\end{array}\right)=36$

No. of ways in which we can select $4$ spots to place the $2$ digits that will appear twice is $=\left(\begin{array}{c}5\\ 4\end{array}\right)=5$

The no. of ways to place the selected $2$ digits at $4$ spots such that doubles can be avoided is $=\frac{4!}{2!2!}=6$

The no. of ways of selecting 5th digit out of the remaining 7 digits is  $=\left(\begin{array}{c}7\\ 1\end{array}\right)=7$

Therefore, the number of ways in which two digits appear twice $=36\times 5\times 6\times 7=7560$

Step 5. Find the no. of 5 -digits numbers that can be formed.

Therefore, the no. of $5$-digits numbers that can be formed are $=15120+30240+7560=52920$