Question

How many 9 digit numbers of different digits can be formed?

Answer 1

Hint: We will use the concept of the relation between permutation and combination. This is given by the formula $P_{r}^{n}=r!C_{r}^{n}$ where $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$. Here n is representing the total number of objects and r is denoting selected items. We will use this relation in order to solve the question.

Complete step-by-step answer:
According to the question we have a total number of digits as 10. This is because we can form any number by using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 which are 10 in total. We are going to find 9 digit numbers with different digits. We will fill up the first place by all the 9 digits and not 10. This is because if zero comes in any 9 digit number and zero is included in it then it will become an 8 digit number instead of 9. Therefore, the first place is filled by 9 digits only or we can say that it will have 9 choices only.
Now the rest of the places can be filled by using the relation of permutation and combination here. We will use the formula $P_{r}^{n}=r!C_{r}^{n}$which results into
$\begin{align}
  & P_{r}^{n}=r!C_{r}^{n} \\
 & \Rightarrow P_{r}^{n}=r!\dfrac{n!}{r!\left( n-r \right)!} \\
 & \Rightarrow P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!} \\
\end{align}$
Here n is representing the total number of objects and r is denoting selected items. As here we have n = 9 and r = 8 therefore, we have
$\begin{align}
  & P_{8}^{9}=\dfrac{9!}{\left( 9-8 \right)!} \\
 & \Rightarrow P_{8}^{9}=\dfrac{9!}{1!} \\
 & \Rightarrow P_{8}^{9}=9! \\
\end{align}$
Therefore, the total number of possibilities is given by $9\times 9!=3265920$.
Hence, 9 digit numbers of different digits can be formed in 3265920 ways.

Note: Alternatively we can use different methods in which we will use the concept of finding out the choices for each place. This is done below.
We can form any number by 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 digits. In order to form a 9 digit number the first digit should be free from 0. Therefore, for the first place we have 9 choices only. So we now have $9,-,-,-,-,-,-,-,-$. We can consider the other places that can be filled by any number also, we can consider 0 for them. But in this question the limitation given to us is that the repetition is not allowed. Therefore, as we have a total 10 digits given by 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 and out of these one is already used in the first place. Therefore, we are left with now 9 choices. That is $9,9,-,-,-,-,-,-,-$. And the third place has 8 choices as there is no repetition. That is $9,9,8,-,-,-,-,-,-$. Continuing in this manner we will get $9,9,8,7,6,5,4,3,2$. Therefore, by multiplying the choices we have a total number of possibilities given by $9\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2=3265920$.
Hence, 9 digit numbers of different digits can be formed in 3265920 ways.