Each hand in the game of bridge has 13 cards dealt from a regular deck of 52 cards.

a) How many different bridge hands are possible?
b) How many different bridge hands have all pour aces in them?
c) How many different bridge hands have no aces in them?

I figured out part A, but am confused about the rest..
52nCr13 = 6.35 x 10^11 hands

a) choose 13 from 52 = C(52,13) = 6.35 x 10^11

you had that

b) you have the four aces, that leaves 9 other cards from the remaining 48
so C(48,9) = 1677106640

c) imagine the deck with no aces, that would be 48 to choose from, but you still want 13 cards
C(48,13) = 1.929x10^11

a) To calculate the total number of different bridge hands possible, we can use the concept of combinations. In bridge, each hand consists of 13 cards from a regular deck of 52 cards.

We can calculate the total number of possible bridge hands using the combination formula, also known as "n choose r" or "nCr." In this case, we have 52 cards and we need to choose 13 cards for each hand. Therefore, the calculation would be written as:

52C13 = 52! / (13!(52-13)!) = 635,013,559,600

So, there are approximately 6.35 x 10^11 different bridge hands possible.

Now, let's move on to the remaining parts:

b) How many different bridge hands have all four aces in them?
To calculate this, we need to consider that all four aces must be included in a hand of 13 cards. We have 4 aces, and we need to choose 9 more cards from the remaining 48 cards (52 - 4 = 48).

So, the calculation would be:

4C4 * 48C9 = 1 * 48! / (9!(48-9)!) = 48! / (9!39!) = 274,068

There are approximately 274,068 different bridge hands that have all four aces in them.

c) How many different bridge hands have no aces in them?
To calculate this, we need to exclude the 4 aces from the deck of 52 cards and choose 13 cards from the remaining 48 cards.

So, the calculation would be:

48C13 = 48! / (13!(48-13)!) = 586,426,440

There are approximately 586,426,440 different bridge hands that have no aces in them.

I hope this clears up any confusion. Let me know if you have any further questions!