How many six−digit numbers are there in all?

Answer:

The smallest digit in the highest place value is 0, which cannot be used. As a result, the highest place value can be 1, the second smallest digit. 

100000 is the smallest six-digit number. 

We all know that the largest digit is nine and can be used everywhere. 

999999 is the largest six-digit number. 

When it comes to AP (Arithmetic progression) 

100000 for the first phrase (a). 

999999 is the last term. 

One is the difference (d). 

The nth term is calculated using the formula. 

a = a + (n – 1)d 

100000 + (n – 1) = 999999 

n – 1 = 899999 

n = 900000 

∴ In total, there are 900,000 6-digit numbers.

OR

The largest 6−digit number is 999999

The smallest 6−digit number is 100000

Hence,  total number of 6−digit numbers =(999999−100000)+1

=899999+1

=900000

=9lakhs

OR

Calculate the total number of 6-digit numbers in all using the permutation idea. First, we’ll determine the number of options for the 6-digit number’s first place digit. The permutation will then be used to determine the number of options available for the remaining digits of the 6-digit number. The total number of 6-digit numbers will be determined by solving this.

It is assumed that the numbers are six digits. 

So, we know that there are only nine options for the first place digits in a six-digit number, i.e. 1,2,3,4,5,6,7,8,9 because 0 cannot be in the first place digit. After all, the number will become a five-digit number. 

As a result, the total number of permutations for the six-digit integer’s initial digit is

= ₉P₁ = 9!/(9−1)! =9

Now we know that there are ten options for the remaining five digits in the six-digit number, namely 0,1,2,3,4,5,6,7,8,9. 

Total number of permutations for the six-digit number’s remaining five digits

= ₁₀P₁=10!/(10−1)! =10

So, the total number of 6 digit numbers in all

=9×10×10×10×10×10=900000

Therefore, there is a total of 9,00,000  6-digit numbers.