Question

How many 8-digit numbers are there in all?

Answer 1

Here, we have to use the concept of permutation for calculating the total number of 8 digit numbers in all. Permutations may be defined as the different ways in which a collection of items can be arranged, in a particular order or sequence. We will find a number of ways to assign different numbers in different places of 8 digit numbers. The arrangement will give us the total number of 8 digit numbers present in total.

Complete step by step solution:
Here, there are 8 different places and we can arrange 10 different number at different position except for the position of the first digit.
We know that for the first place digit in the 8 digit number there are only 9 choices available i.e. from 1 to 9. We cannot put 0 in the first place digit because then it will become a 7 digit number.
Therefore, total number of permutation for the first digit in the 8 digit number $= {}^9{{\rm{P}}_1} = \dfrac{{9!}}{{{\rm{(9}} - {\rm{1)}}!}} = 9$
Now, we know that for the rest of the 7 digits in the 8 digit number there are 10 choices available that are from 0 to 9.
Therefore, total number of permutation for all the remaining 7 digits in the 8 digit number $= {}^{10}{{\rm{P}}_1} = \dfrac{{10!}}{{{\rm{(10}} - {\rm{1)}}!}} = 10$
So, total number of 8 digit number in all $= 9 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 90000000$

Hence, there are a total of 90,000,000 8-digit numbers.

Note:
We can solve the given question using the following formula:
Total 8 digit number$=$ (Largest 8 digit number $-$ Smallest 8 digit number)$+ 1$
Total 8 digit number $= \left( {99,999,999 - 10,000,000} \right) + 1 = 90000000$