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For the question

How many odd three-digit numbers can be written using digits from the set $\{2,3,4,5,6\}$ if no digit may be used more than once?

I thought that the answer would be $2\choose1$ to pick the odd number at the end and that leaves 4 numbers from which to choose the remaining two spots. So the answer is $5\cdot 4 \cdot {2\choose1} = 40$, but the answer is $4\cdot 3 \cdot {2\choose1} = 24$ and I don't understand why the remaining odd number is not counted?

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2 Answers 2

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Because the "remaining" odd number is the number you have already used (as the last digit) so for the other two digits you are choosing among $4$ possibilities, not $5$.

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You have calculated the answer with replacement of the chosen odd digit for the end while the question states without replacement. When we take a digit (whether $3$ or $5$) we are left with $4$ more digits not $5$.

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