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Saumyojit

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how many ten-digit numbers have at least two equal digits

First digit has 9 options except 0 .
Now to ensure two equal digit I fix any one out of 9 places --> 9c1
Now in the remaining places I have 10 options -> 10^ 8

9* 9c1 * 10^8

Where am i wrong
 
Saumyojit said:
how many ten-digit numbers have at least two equal digits

First digit has 9 options except 0 .
Now to ensure two equal digit I fix any one out of 9 places --> 9c1
Now in the remaining places I have 10 options -> 10^ 8

9* 9c1 * 10^8

Where am i wrong
Click to expand...

First, I would start by counting numbers that have no equal digits. It's standard to solve "at least" problems using the complement.

But as for your method, I don't see that it has anything to do with the question! How does "fixing" one place ensure two equal digits? Do you mean that's the one place that has the same digit as the first, or something like that?
 
Dr.Peterson said:
Do you mean that's the one place that has the same digit as the first,
Click to expand...
Yes.
First i select what is the first digit (9ways) .Assume 7 then out of nine remaining places there will be any one place which will be filled by 7 (9c1) , (7_ _ _ _ _ _ _ _7) then in remaining eight places there will be 10 ^ 8 options . So , eg : 7555321077, 7000000007 , 7555700044

First it will fill the leftmost first place of number , then assume one blank is not there , so Eight blanks are there . So it will list all the numbers starting with that digit and a similar digit which is present in any of the 9 blanks remaining so that atleast two is found.

This will happen for all 9 digit. Each DIGIT each List

What is wrong?
 
Last edited:
Saumyojit said:
Yes.
First i select what is the first digit (9ways) .Assume 7 then out of nine remaining places there will be any one place which will be filled by 7 (9c1) , (7_ _ _ _ _ _ _ _7) then in remaining eight places there will be 10 ^ 8 options . So , eg : 7555321077, 7000000007 , 7555700044

First it will fill the leftmost first place of number , then assume one blank is not there , so Eight blanks are there . So it will list all the numbers starting with that digit and a similar digit which is present in any of the 9 blanks remaining so that at least two is found.

This will happen for all 9 digit. Each DIGIT each List

What is wrong?
Click to expand...
You are counting the same number more than once. For example, your 7555321077 would also be obtained as 7555321077.

You are also missing numbers for which some other digit than the first is repeated, such as 7555321049.

When you think of a way to count, you must always then think about whether you might be over- or under-counting. This is why I always ask you to describe the method, not just show a calculation: I hope that in writing, you will also be thinking!
 
Saumyojit said:
Yes this way it is 9* (10^9) - 9*9!
Click to expand...
That is the way I would count the ten digit numbers with no repartitions.
There are nine ways to pick the leading non-zero digit. That leaves nine other digits to fill out the other nine places.
Thus there are [imath]9\cdot 9!=3265920[/imath] ten digit numbers having no digit repeated.
So there are [imath]9\cdot 10^9 -9\cdot 9!=8,996,734,080[/imath] ten digit numbers having at least two digits repeated.
 
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