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2. How many codes of 4 different numbers are there if the first digit has to be a 1 or a 2 or 3?

User RazorHead
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Given a 4-digit number, with all of them different from each other, if the first digit must be 1, 2, or 3, then the last 3 numbers can not be 123, 132, 213, 231, 312, or 321. Then, we have 10 numbers to choose from, from 0 to 9, to the next 3 numbers, excluding the cases listed before (6 in total).

Using the combination operator, the number of different ways to choose 3 numbers out of 9 (because the first one is reserved) is:


C^9_3=(9!)/(3!\cdot6!)=84

Excluding the 6 cases listed above, this leads to a total of 78 combinations. Finally, since there are 3 options for the first digit, we multiply this result by 3, leading to 234 different ways.

User Onoria
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