In a game of bridge with a deck of 52 cards, Determine the probability of having a hand containing exactly 4 kings, 4 queens, and 5 hearts?

Question

In my attempt i'm not sure if i am under counting by having 13C2 instead of 13C1 and 12C1. Thanks.

My attempt:

[(13C2)(4C4)(4C4)(11C5)(4C1)]/52C13.

Answer 1

As you have correctly determined, the denominator must be ${}_{52}C_{13}$. In the numerator, we need the total number of hands that satisfy the requirement of having 4 kings, 4 queens, and 5 hearts.

In this case, the task is easier than you seem to be making it. In particular, there are exactly 4 kings and 4 queens in the deck, which means that the number of ways to select these 8 cards is $1$.

Now, we need the number of ways in which we can select the $5$ hearts that form the remainder of the hand. Since we have removed the kings and queens, there are $11$ hearts to choose from. Thus, the total number of ways that we can select $5$ such hearts is $_{11}C_{5}$.

Thus, number in the numerator should be $1 \cdot {}_{11}C_5 = {}_{11}C_5.$