probability without combinations I have a problem in solving a probability sum. I have to find probability of 2 cards from diamond and another 2 cards from heart from 52 deck of cards. 
They solved it using combinations like, $$\frac{13C2 \cdot 13C2}{52C4}$$
But I want to solve it like 
$$\frac{26}{52} \cdot \frac{25}{51} \cdot \frac{24}{50} \cdot \frac{23}{49}$$ 
But they are unlike, why?
 A: The given method does not work because there are two different sets and there must be two cards of any of the sets.
The method would be correct if the requirement would be : $4$ red cards.
You could solve the exercise with a tree-diagramm, but this will be more difficult because we have to consider some paths.
A: Any first red card would be right: $\frac{26}{52}$.
The second, third, or fourth card (introducing a factor of 3) could be the other red card of the same suit as the first card with chance  $\frac{12}{51}$.
The other two cards are from the different red suit with  $\frac{13}{50}$. and  $\frac{12}{49}$.
This would get:  $$\frac{26 *3 *12 * 13 *12}{52 * 51 * 50 *49}$$
A: $$\dfrac{26}{52} \cdot \dfrac{25}{51} \cdot \dfrac{24}{50} \cdot \dfrac{23}{49}
 \;=\; \dfrac{^{26}\mathrm C_4}{^{52}\mathrm C_4}$$
This is the probability of selecting any 4 of 26 cards (red) out of the deck of 52 cards.
However you wanted to the probability of selecting 2 of the 13 diamonds and 2 of 13 the hearts out of the deck.
$$\dfrac{13}{52} \cdot \dfrac{12}{51} \cdot \dfrac{13}{50} \cdot \dfrac{12}{49}
 \;=\; \dfrac{{^{13}\mathrm C_2}\cdot{^{13}\mathrm C_2}}{^{52}\mathrm C_4}$$

Or alternatively: pick any red card, then either another from the same suit and two from the other, or one from the other suit, one from any suit and one from the other of that, but as order of the suits doesn't actually matter, we divide by the number of permutations.
$$\dfrac{26}{52}\cdot\Big(\dfrac{12}{51}\cdot\dfrac{13}{50}\cdot\dfrac{12}{49}+\dfrac{13}{51}\cdot\dfrac{24}{50}\cdot\dfrac{12}{49}\Big)\times\dfrac{2!\;2!}{4!} \;=\; \dfrac{{^{13}\mathrm C_2}\cdot{^{13}\mathrm C_2}}{^{52}\mathrm C_4}$$
In the end, combinations is the best tool for the job.
