Probability of drawing two cards such that one is a spade and other one is a heart.

Question

I got the answer of this question via two methods

$(13÷52)*(13÷51)*2............(i)$

And the second is

$(13C1*13C1)÷(52C2).........(ii)$

I know that both will give same numerical correct answer, but I'm confused from the fact that in first equation we multiplied by $2$ which means that either spade or heart can be drawn first so why can't we multiply second equation using same logic? Please help if I'm missing something really important. Thanks in advance.

Answer 1

For the first formula, order is taken into account. We are counting ordered pairs of cards. If you don't multiply by $2$ you'll miss the drawing in the opposite order.

For the second one we are counting sets, where order does not matter. Exchanging the two elements of the set changes nothing, so if you multiply by two you'll get the wrong answer.

Answer 2

Let me suggest another way to solve the problem that might help you think about it. You have $26$ choices out of $52$ of a first card that keeps you "alive." After choosing that first card, you have $13$ choices out of $51$ of a second card that keeps you alive.