I have this problem
Question: Four cards are drawn at random from a pack of 52 cards. Find the probability that two are kings and two are queens.
Answer: $$ P(E) = \frac{\,^4C_2\cdot \,^4C_2}{\,^{52}C_4} $$
That's what the answer should be but I wanted to verify it so, I started to draw a tree which represents all the outcomes.
For example:
We can pick any of the 4 kings let's call them $k_1, k_2, k_3, k_4$ and we have $4$ queens call them $q_1, q_2, q_3, q_4$.
In first case we pick $k_1$, now we can pick $k_2, k_3, k_4$. say we pick $k_4$. Now we can pick $q_1, q_2, q_3, q_4$.
We pick $q_1$, now we can pick $q_2, q_3, q_4$
If we look at number of outcomes this way we get :- $3\cdot(3\cdot 4)\cdot 4 = 144$
This is different from $^4C_2\cdot\,^4C_2=36$
Why has this difference in answers arisen? Am I right or wrong? And if I am wrong, where have I gone wrong here?
As @lulu pointed out in the comment, you have double counted the ways in which two kings(similar for two queens) can be drawn.
All that matters here is drawing two kings (and queens) and not the order in which you draw them, so drawing $K_1$ first and then $K_2$ is the same as drawing $K_2$ first and then $K_1$ but you have counted both the possibilities separately and you have done this for all possible pairs in the set of kings (similarly for the set of queens).
So you must divide the answer you got by a factor of $4$ to overcome the double counting in both the set of cards.