Question 429255
Your first "solution" actually suggested the number of ways to pick three of *any* card out of a deck of 52 cards, where the order does matter.


The second solution is closer, you picked three aces out of four cards, using the expression 4P3. However this over-counts quite a bit, because 4P3 assumes the order also matters. Since we are choosing three aces and the order is assumed not to matter, it would be represented as 4C3, or 4!/(3!1!), which is equal to 4.