Question

How many numbers are there between 100 and 1000 in which all the digits are distinct?

Answer 1

Hint: We will apply the concept of permutation here. The formula of permutation is given by $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ where n are the total number of objects and r is the selected objects. We will use this formula to find specially the arrangements made by the digits.

Complete step-by-step answer:
Now first we will understand the language of the question which says that there are two numbers 100 and 1000 that he has considered. And we are supposed to find out how many numbers lie between 100 and 1000. As we know that any that will lie between 100 and 1000 will be a three digit number. So, we will form places like so $-,-,-$.
Now we need focus here. We can form any number by 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 digits. In order to form a three digit number first digit should be free from 0. Therefore, for the hundredth place we have 9 choices only. So we now have $9,-,-$. Now we will apply permutation for the other two places. After filling up the hundredth place we are left with 9 choices and we need to fill the rest of the places which are 2. Therefore, we have n = 9 and r = 2.
By the formula of permutation is given by $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ where n are the total number of objects and r is the selected objects. We have
$\begin{align}
  & P_{2}^{9}=\dfrac{9!}{\left( 9-2 \right)!} \\
 & \Rightarrow P_{2}^{9}=\dfrac{9!}{7!} \\
 & \Rightarrow P_{2}^{9}=\dfrac{9\times 8\times 7!}{7!} \\
 & \Rightarrow P_{2}^{9}=\dfrac{9\times 8\times 1}{1} \\
 & \Rightarrow P_{2}^{9}=9\times 8 \\
 & \Rightarrow P_{2}^{9}=72 \\
\end{align}$
Now the other two places can be filled by $P_{2}^{9}=72$ ways. Therefore, the total number of possibilities here is given by
$\begin{align}
  & 9\times P_{2}^{9}=9\times 72 \\
 & \Rightarrow 9\times P_{2}^{9}=648 \\
\end{align}$
Hence the total number of possibilities is 648.

Note: Alternatively we can use a different method which is filling up method to fill the tenth and ones place. In this method we will use the concept of finding out the choices each place can have. 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 three digit number first digit should be free from 0. Therefore, for the hundredth place we have 9 choices only. So we now have $9,-,-$. We can consider the other two places 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 hundredth place. Therefore, we are left with now 9 choices. That is $9,9,-$. And the one place has 8 choices as there is no repetition. That is $9,9,8$. Therefore, by multiplying the choices we have a total number of possibilities given by $9\times 9\times 8=648$.
Hence the total number of possibilities is 648.