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How many numbers are there between 100 and 1000 in which all the digits are distinct?

My mathematics textbook says the answer is 648.

My analysis:-

The number at the hundred's place can be chosen in 9 ways(as zero is not possible)

The number at the ten's place can be chosen in 9 ways(the number at the hundred's place is no longer available but zero is now available).

The number at the one's place can be chosen in 8 ways(the number at the hundred's and ten's place is not available but zero is available).

So by the multiplication rule, the total number of ways would seem to be 9*9*8=648.

But the hundred's place can have 1 at its place, ten's can have 0 and ones can have 0. So 100 is a possibility but the question is asking for numbers between 100 and 1000.As a result of that, 100 has to be removed. Therefore, the correct answer is 648-1=647.

I have checked and rechecked my method and can't find anything wrong with it. Can someone please help verify my answer?

Abhishek Mhatre
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2 Answers2

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The answer $648$ is correct.

There are nine choices for the hundreds place. There are also nine choices for the tens place (it can be any digit except what you chose for the hundreds).

Finally, there are eight choices for the ones place, regardless of what the tens place is. If the tens place is a zero, then you can't choose either the hundreds place digit or zero for the ones place. If the tens digit is not zero, then you can pick zero for the ones place, just not the digits in the hundreds or tens place.

John
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  • What do you mean by "a case that wasn't present in the first place"? Of course the 1 is present in the first operation. The thing is that the last zero is sorted out in the thirs operation. – amsmath Oct 31 '17 at 17:16
  • @amsmath I edited my answer. – John Oct 31 '17 at 17:20
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The digits in 100 are not distinct. So, 100 is sorted out by your algorithm.

amsmath
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