# 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.

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 $$k,k,q_2,q_1$$)

• You count three kings one queen three times, so need to subtract them twice. Sep 20 '20 at 4:45
• Yes thats right. Sep 20 '20 at 4:59

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.