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Given all four digits odd numbers.How many of them have exactly two even numbers?

The sample set is $ \{0,1,2,3,4,5,6,7,8,9\}$. To have a odd number, the last digit have to be odd, so there are five possible ways to the last digit. In order to have a four digit number, the first can't be zero. In a four digit number, if exactly two are even, the other two have to be odd( one of them is the last). So, I made a permutation with replacement regarding to the five odd numbers and two positions.The result was $5^2=25$.

In other hand, the total of four digit odd numbers is $9\cdot 10^2 \cdot 5$. Nine for the possible ways to the first digit, $10^2$ for the second and third digit and $5$ to the last.

At last I subtracted $25$ from $4500$.However the result don't match to the solutions.Can you help me? Thanks

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    Your solution starts correctly, but I don't see what you mean with "So, I made a permutation with replacement ... The result was 25". It should be clear that there are more than 25 numbers you should substract from all the 4-digit numbers that end with an odd digit to get what you want.2012-01-16

2 Answers 2