How many 4 digit even numbers have all 4 digits distinct
Answers
On problems like this one, it is best to start counting with what has the most constraint. Here we need the last one even, and the first one to be non-zero. The two central ones don't have constraint aside being distint of everyone else.
We start by dividing the counting whether the last digit is 0 or not.
If the number ends with a 0 then there are 9 choices for the first digit, 8 for the second and 7 for the third, which makes 1×9×8×7=504 possibilities.
If the number is even ending with something else than 0 then there are 4 choices for the last digit, 8 choices for the first digit (no 0 nor the last digit), 8 for the second digit and 7 for the third digit, which makes 4×8×8×7=1792
Together, this gives 2296 numbers with 4 distinct digits that are even. Note that this does not allow leading 0, as you see to want it based from the question.
Hope this helps.
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We start by dividing the counting whether the last digit is or not.
If the number ends with a then there are choices for the first digit, for the second and for the third, which makes possibilities.
If the number is even ending with something else than then there are choices for the last digit, choices for the first digit (no nor the last digit), for the second digit and for the third digit, which makes
Together, this gives numbers with distinct digits that are even. Note that this does not allow leading , as you see to want it based from the question.
extra
it is best to start counting with what has the most constraint. Here we need the last one even, and the first one to be non-zero. The two central ones don't have constraint aside being distint of everyone else.