There are 5 numbers missing and they are 1, 2, 5, 7, 8
Once you choose the first one (out of five), there are four left to choose from
Once you choose the second one (out of the remaining four), there are three left to choose from
Once you choose the third one (out of the remaining three), there are two left to choose from
Once you choose the fourth one (out of the remaining two), there is only one left
For example, if you first choose the 7 then remaining is 1, 2, 5, 8
Then if you choose 1, then remaining is 2,5,8
Then if you choose the 8, remaining is 2 and 5
Then if you choose the 2, the 5 is left over by default.
So the choices are: 5 x 4 x 3 x 2 x 1 = 120 different numbers that can be possible answers to
"a nine digit non-repeating number that ends in 6349"
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