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Number of $4$ digit numbers with no repeated digit is

  1. $4536$
  2. $3024$
  3. $5040$
  4. $4823$

Well, I am very much weak in combinatorics. Please help.

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up vote 6 down vote accepted

Ok so lets write down any old $4$ digit number


How many choices do we have for the digit $a$? We have $9$ choices (since the first digit cannot be $0$). Now for each possible choice of $a$ we have $9$ choices for $b$ (since we want $b$ to be a different digit to $a$ and we now allow $0$).

So for choice of the $ab$ part we have $9*9 = 81$ possibilities.

Now for each of these we have $8$ choices for $c$ (to avoid $c$ being the same as either $a$ or $b$). And for each of these we have $7$ choices for $d$ (to avoid $d$ being the same as either $a,b$ or $c$).

So in total there are $9*9*8*7 = 4536$ possible numbers.

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ah! so beautiful answer thank you fretty. – La Belle Noiseuse Jun 11 '12 at 11:45
I can't help but notice that there are no repeated digits in the number $4536$. Somehow the answer to the question was among the numbers being counted, how mysterious! – Marc van Leeuwen Jun 11 '12 at 13:11

A bit more hint: Ask yourself how many ways can you choose the digit in the first position, then how many ways can you choose the second-place digit ...

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Hint: this is like drawing four digits out of a bag of ten digits 0-9. How many different ways can this happen?

Edit: To correct the model I had in mind, the bag would first have to contain only 1-9. After you've picked the leftmost digit, you would throw in a 0 ball and continue to pick the last three digits.

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Not exactly, since the choice of the first digit is restricted. – fretty Jun 11 '12 at 11:37
@fretty True! A scurvily written question. – rschwieb Jun 11 '12 at 11:42

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