A standard 52-card deck contains four queens, twelve face cards, thirteen hearts (all red), thirteen diamonds (all red), thirteen spades (all black), and thirteen clubs (all black). Of the 752,538,150 different eight-card hands possible, decide how many would consist of the following.
(a) all clubs
(b) all black cards
(c) all queens
(a) There are $◻$ ways to have a hand with all clubs.
(Simplify your answer.)

Step 1 ：A standard 52-card deck contains four queens, twelve face cards, thirteen hearts (all red), thirteen diamonds (all red), thirteen spades (all black), and thirteen clubs (all black). Of the 752,538,150 different eight-card hands possible, decide how many would consist of the following.

Step 2 ：(a) all clubs

Step 3 ：To solve this problem, we need to use the concept of combinations in probability. A combination is a selection of items where order does not matter. In this case, we are selecting 8 cards out of 13 clubs. This can be calculated using the combination formula: $C(n,k)=\frac{n!}{k!(n-k)!}$, where n is the total number of items, k is the number of items to choose, and '!' denotes factorial.

Step 4 ：Let n = 13 and k = 8

Step 5 ：Using the combination formula, we find that there are $C(13,8)=1287$ ways to have a hand with all clubs.

Step 6 ：Final Answer: There are $\boxed{{\displaystyle 1287}}$ ways to have a hand with all clubs.