Question

The total number of 9 – digit numbers of different digits is
(a) \[10\left( 9! \right)\]
(b) \[8\left( 9! \right)\]
(c) \[9\left( 9! \right)\]
(d) None of these.

Answer 1

Hint: We solve this problem by using the permutations. We have 10 digits from 0 to 9. We need to place these 10 digits in nine places to get a 9 – digit number keeping in mind that the first digit should not be zero because a number does not start with 0.

Complete step-by-step answer:
We are asked to find the number of 9 – digit numbers of different digits.
Let us assume that there are 9 boxes representing 9 places of 9 – digit numbers.

We know that there are 10 digits ranging from 0 to 9.
We know that any number does not start with 0.
So, we can say that the first box have 9 possibilities
We are given that the digit should not be repeated. So, the second place has 9 possibilities assuming one digit has already occupied first place and 0 can be placed in second place.
Now, we can say that the third place has 8 possibilities.
Similarly, we can say that fourth place has 7 possibilities and so on.
Now, let us write the number of possibilities in each box as follows

Now that total number of possibilities is found by multiplying all the possibilities as follows
 \[\begin{align}
  & \Rightarrow N=9\left( 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2 \right) \\
 & \Rightarrow N=9\left( 9! \right) \\
\end{align}\]
Therefore, we can say that there are a total of \[9\left( 9! \right)\]number of 9 – digit numbers of different digits.
So, option (c) is the correct answer.

So, the correct answer is “Option (c)”.

Note: Students will make mistakes in taking the possibilities of first place. There will be no number that starts with 0. If the 9 – digit number starts with 0 then it becomes 8 – digit number which is not required in this case. So, taking the total possibilities in the first place is very important in this problem. Only this part should be taken care of.