How many different lists $a, b, c, d$ of distinct odd positive integers with $a<b<c<d$ have the property that $a+b+c+d=24$ ?
There are exactly 5 lists with this property, namely

$$
3+5+7+9=1+5+7+11=1+3+5+15=1+3+7+13=1+3+9+11=24
$$

Why are these the only lists?

If $a=3$, then since $a, b, c, d$ are odd positive integers with $a2 c$ and so $c<9$.

Since $c$ and $d$ are odd positive integers, then $c=7$ and so $d=11$.

If $b=3$, then $c+d=24-1-3=20$.

Therefore, using a similar argument, we get $c<10$ and so $c=9$ (which gives $d=11$ ) or $c=7$ (which gives $d=13$ ) or $c=5$ (which gives $d=15$ ).

Therefore, there are exactly 5 lists with the desired properties.

ANSWER: 5