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How many 3 digit numbers can be formed from the digits 0-9 if the digits cannot be repeated?

Guest Apr 23, 2017
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These confuse me too.. but let's just start writing out all the possibilities and then maybe we can see what the answer will be.

First digit 0:

01       02       03       04       05      06      07     08     09

012     021     031     041      There will also be exactly

013     023     032     042      eight possibilities under each

014     024     034     043      of these.

015     025     035     045

016     026     036     046

017     027     037     047

018     028     038     048

019     029     039     049

Iff the first digit is 0, there are 8 * 9 possibilities.

If the first digit is 1, there are also 8 * 9 possibilities.

There are 10 possible first digits, so there are 10 * (8 * 9) = 720 possibilities.

You can also just use this formula: $$_nP_r=\frac{n!}{(n-r)!}$$

Where n = 10 and r = 3

Here's a really good explanation of that formula:

https://www.youtube.com/watch?v=DROZVHObeko

hectictar  Apr 23, 2017

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