In how many ways can 4 cards be drawn randomly from a pack of 52 cards such that there are at least 2 kings and at least 1 queen among them? So i tried this question in two ways
(i)In my first method I made different possible arrangements and then find the number of ways
So, the different possibilities are:
2 kings and 1 queen and 1 other card
Or, 3 kings and 1 queen
Or, 2 kings and 2 queens
Total ways possible = $${4\choose2}{4\choose1}{44\choose1}+{4\choose3}{4\choose1}+{4\choose2}{4\choose2} = 1056 + 16 + 36
= 1108
$$
Total ways possible = 1108
And this is the correct answer.
(ii) In order to shorten the above method, I did this
We need at least 2 kings and 1 queen, so total ways possible = $${4\choose2}{4\choose1}{49\choose1}=1176$$(49 because I subtracted the 3 cards from the deck of 52 cards).
So what's the problem with second method? Why I'm getting additional 68 ways (1176 - 1108= 68)? And Is there any way to solve this question without making cases?
Thanks and stay safe.
 A: You're overcounting in the second method.
Label the four kings $K_1, K_2, K_3, K_4$. In the second method, you will count the case in which you pick an arbitrary queen $Q$ followed by $K_i$ followed by $K_j$ separate from the case in which you pick a queen $Q$ followed by $K_j$ followed by $K_i$. However, these two outcomes should be treated as the same outcome.
A: Since in your second method, for example you have the case 3 kings 1 queen counted three times and 2 kings 2 queens counted twice.
Since assume that ${4\choose2}$ picked up the king of spades and the king of hearts, and then the ${49\choose1}$ picked up the king of diamonds. $k_s,k_h,q,k_d$ this is one way of choosing your cards. But this method counts $k_d,k_h,q,k_s$ as another way and this is wrong they should only be counted once.
On the other hand, in your first method when you use ${4\choose3}$ for the case where there are 3 kings, you're being sure that all the equivalent ways are counted once.
Now why they are 68?
${4\choose3}×2×4 + {4\choose 2}×{4\choose 2}=32+36=68$
${4\choose 3}$ for the case we have 3 kings, $2$, because we want to cancel 2 ways (i.e. if we have $k_1,k_2,q,k_3$ we want to cancel $k_1,k_3,q,k_2$ and $k_3,k_2,q,k_1$ since they must be counted only once) and the $4$ because we have 4 possibilities for the queen.
Simlarly for the queens, ${4\choose 2}$ for the case we have 2 queens,  and another ${4\choose 2}$ because we have ${4\choose 2}=6$ possibilities of the first 2 kings. (Note that here we don't multiply by $2$ here since if we have $k,k,q_1,q_2$ we only want to cancel only 1 equivalent case and that is $k,k,q_2,q_1$)
