How many different $7-$ digit positive integers exist? (Note that we do not allow $''7-$digit$''$ integers that start with $0,$ such as $0123456;$ this is actually a $6-$digit integer.)
Answer
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Hint: We will not consider the integer that starts with $0$ a $7-$ digit for it is a $6-$digit number. We will use the rest of the numbers from $1$ to $9$ in the first place. Then, $0$ can be included in any of the other places.
Complete step by step solution:
We are asked to find the number of $7-$digit positive integers. We have the digits that range from $0$ to $9.$ From these numbers, we will only consider nine digits from $1$ to $9$ to put in the first place for the integers starting with $0$ and contain six other digits are $6-$digit numbers.
So, we have the digits $1,2,3,4,5,6,7,8$ and $9$ to put in the first place.
So, as we know, any of these $9$ numbers can be the first digit.
Since we are asked to find the number of all $7-$digit positive integers exist in the number system, the repetition is allowed.
So, we can put any of the ten digits from $0$ to $9$ in the second place. So, we can possibly accommodate any of the $10$ digits in the second place.
We can repeat the same with next places also.
Because repetition is allowed, we can put any of the $10$ digits in the third, fourth, fifth, sixth and seventh places.
So, we will get $9\times 10\times 10\times 10\times 10\times 10\times 10$ integers.
Hence there are $9000000$ $7-$digit positive integers.
Note: We know that if repetition is not allowed, then there are $9\times 9\times 8\times 7\times 6\times 5\times 4$ $7-$digit positive integers in the number system with distinct integers in the distinct places. But when the repetition is allowed, we always have the choice of choosing any of the total numbers for each place.
Complete step by step solution:
We are asked to find the number of $7-$digit positive integers. We have the digits that range from $0$ to $9.$ From these numbers, we will only consider nine digits from $1$ to $9$ to put in the first place for the integers starting with $0$ and contain six other digits are $6-$digit numbers.
So, we have the digits $1,2,3,4,5,6,7,8$ and $9$ to put in the first place.
So, as we know, any of these $9$ numbers can be the first digit.
Since we are asked to find the number of all $7-$digit positive integers exist in the number system, the repetition is allowed.
So, we can put any of the ten digits from $0$ to $9$ in the second place. So, we can possibly accommodate any of the $10$ digits in the second place.
We can repeat the same with next places also.
Because repetition is allowed, we can put any of the $10$ digits in the third, fourth, fifth, sixth and seventh places.
So, we will get $9\times 10\times 10\times 10\times 10\times 10\times 10$ integers.
Hence there are $9000000$ $7-$digit positive integers.
Note: We know that if repetition is not allowed, then there are $9\times 9\times 8\times 7\times 6\times 5\times 4$ $7-$digit positive integers in the number system with distinct integers in the distinct places. But when the repetition is allowed, we always have the choice of choosing any of the total numbers for each place.
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