How come lining up only some of the things still involves factorials?

Ten runners, three different medals. The first medal has ten candidates. Once that runner is chosen, the second has nine. Once both are chosen, the third has eight. The product is $10\times9\times8=720$. It stops after three factors because there are only three places to fill. A full line-up of all ten would be $10!=10\times9\times8\times7\times6\times5\times4\times3\times2\times1=3{,}628{,}800$. That larger number counts an order for every runner, including the seven who won nothing.

The stopping product is not a different rule. It is the multiplication rule with the options running out at three places instead of ten. The first place has ten options, the second has nine, the third has eight, and there is no fourth place, so the next factor is never multiplied. The number of factors equals the number of places filled, so a partial line-up is a factorial cut short rather than a factorial altered. This is why the answer is smaller than $10!$ but still built from the same descending factors. Nothing was divided yet; the product simply stopped.

The tempting repair is to take $10!$ and divide by $3!$, because three medals were awarded. That gives $10!/3!=3{,}628{,}800/6=604{,}800$, which is not $720$. Dividing by $3!$ would be the repair for three indistinguishable medals, not for leaving seven runners out. The seven non-medalists are not arranged as first, second and third leftovers. They are simply the rest of the field, with no order attached.

For every medal result, the seven remaining runners can be ordered in $7!$ ways. Since $7!=7\times6\times5\times4\times3\times2\times1=5{,}040$, each genuine medal line-up is counted $5{,}040$ times by the full factorial. Dividing by $7!$ collapses those copies back to one. The quotient is not a correction invented to make the answer smaller; it is the exact overcount. The denominator is the size of the group whose internal order you decided not to care about. That group is the leftovers, not the medal places.

Take one medal result: A first, B second, C third. The full factorial counts A, B, C, D, E, F, G, H, I, J and A, B, C, J, I, H, G, F, E, D as different, because the last seven are in different orders. Both give the same three medals. For each medal result there are $5{,}040$ such tail orders, so the full count is $5{,}040$ times too large. Divide and the tail disappears.

The top factorial is the full pile. The bottom factorial is the untouched pile. Stopping the product before the untouched pile is the same as dividing by the number of orders inside that pile. If you fill five of ten places, the product is $10\times9\times8\times7\times6=30{,}240$, and the quotient is $10!/5!=3{,}628{,}800/120=30{,}240$, because the five leftovers have $5!$ internal orders. The shape is always the same: full factorial over leftover factorial, with the number of filled places deciding where the tail is cut.

If all ten are lined up, the untouched pile is empty and $0!=1$, so the quotient is $10!/0!=10!$, as it should be. If the seven leftovers did have to be ordered, then the problem is a full line-up of ten and the answer is $10!$, not $10!/7!$. If some of the three chosen are identical, a further division is needed for their swaps. The partial factorial rule assumes the chosen items are distinct and that their order matters; if either assumption fails, the denominator changes.