# 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?

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?

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 $$5$$.
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 $$5$$.