Question

How many $6$ digit numbers can be formed by using the digits $0,1,2,3,4,5?$

Answer 1

If a number has zero in the leftmost side, then that number has a number of digits one less than the total count. So, if a number contains six digits with zero in the leftmost side, then that number is a five-digit number. We will try the remaining digits in the leftmost side.

Complete step by step answer:
Let us consider the given six digits. They are $0,1,2,3,4$ and $5.$
Now, we are asked to find the number of six-digit numbers that can be formed by using the given six digits.
A six-digit number does not contain zero in the leftmost side. So, we can say that the numbers we need to count do not include the numbers with zero in the leftmost side.
This clearly shows that the digits that can occupy the leftmost side are $1,2,3,4$ and $5.$
Suppose that these digits are not repeated.
We can say that there are only $5$ possible digits among the given six digits that can be at the first position from the left.
We know that even if one of these digits is taken to the first position from the left, any of the remaining $5$ digits can be put in the second position. Then, the number of possible digits for the third position is $4,$ for the fourth position is $3,$ for the fifth position is $2$ and for the sixth position is $1.$
Therefore, the number of six-digit numbers that can be formed by the given numbers is $5\times 5\times 4\times 3\times 2\times 1=600.$
Now, suppose that the digits are repeated. Then, at the first position from the left, we can accommodate $5$ digits. Since zero cannot be put at this position. We can accommodate all the six digits given in the remaining positions.
Therefore, the number of six-digit numbers that can be made by the given digits is $5\times 6\times 6\times 6\times 6\times 6=38,800.$

Hence the number of six-digit numbers that can be made by using the given digits when the digits cannot be repeated is $600$ and when the digits can be repeated is $38,800.$

Note: When we have six digits, we can make $6!$ six-digit numbers. But when zero is put at the first position, the numbers made are five-digit numbers. So, we can make $5!$ five-digit numbers. So, we need to subtract these numbers from the first found $6!$ numbers. So, we will get $6!-5!=6\times 5!-5!=\left( 6-1 \right)5!=5\times 5!=600$ numbers when the digits are not repeated.