A simple combinatorics problem

dimanche 31 août 2014

1. The problem statement, all variables and given/known data

How many 4-digit odd numbers are there? (Without repetition)





2. Relevant equations







3. The attempt at a solution

I thought in the following way: For the last digit, we have 5 possibilities(1,3,5,7,9). For the last second we are left with 9 possibilities, 8 for the next one and 6 for the next one( because 0 is not allowed). So, I got 9 x 5 x 8 x 6 = 2160. But the answer is 2240.



Another approach: 5 for the 4th one, 8 for the 1st, 8 for the 2nd and 7 for the 3rd which gives the correct answer.



But I don't understand why the first method gave the wrong answer. I guess it is something like when the first 3 digits are odd, the last one has only 2 ways of filling, but I am not sure.





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