When doing at least/at most problems in combinations, why must all possibilities be accounted for separately? For example, in a standard deck of playing cards a hand of 5 cards is chosen, at least three of them must be red. So first I know that there are 26 red cards and I can choose 3 of them. My initial thought was that, since the other 2 cards can be any colour, there are 49 cards to choose from (52 minus the three cards I have chosen) and I would choose 2 of them go complete the hand. However this is not the right way to solve the problem, apparently I need to account for each possibility separately, so I need to find the possibilities for 3 red and 2 black cards, plus 4 red cards and 1 black card, plus 5 red cards and 0 black cards. I understand that these two methods of solving get different answers but I'm confused about why that is and why I need to account for each possibility separately instead of just allowing the last 2 cards to be any unselected card. 
 A: The problem is that you risk overcounting if you are not careful.  For example, consider the hand $$\{A\heartsuit, K\heartsuit, Q\heartsuit, J\heartsuit, 10\spadesuit\}$$
How would you account for this hand?  You could say:  Choose the triple $\{A\heartsuit, K\heartsuit, Q\heartsuit\}$ and then pick $\{J\heartsuit, 10\spadesuit\}$ as your last two cards.  Or you might choose the triple $\{A\heartsuit, K\heartsuit, J\heartsuit\}$ and then pick $\{Q\heartsuit, 10\spadesuit\}$ as your last two cards. Thus you would count this hand more than once.  In fact, you'd end up counting it four times (once for each red card). Similarly, you'd count the hand $$\{A\heartsuit, K\heartsuit, Q\heartsuit, J\heartsuit, 10\heartsuit\}$$ $\binom 52=10$ times, once for each pair of red cards in the hand.
For clarity, count the pairs of cards that contain at least one red card.  If you are careless about the "extra" card you'd get $26\times 51=1326$.  To get the correct answer, first count the pairs with exactly one red card (that gives $26\times 26= 676$ as you need to choose one of each) and then add the pairs with two red cards (that gives $\binom {26}2=325$).  the total is then the sum $676+325=1001$ so you can see that the first method badly overcounts.  Note that, as expected, you are counting each of the pairs with two red cards twice so in this case the error is exactly $\binom {26}2=325$.  In general, the error is not so easy to count.
