Counting Homework

Question

How many different 5 card hands can be dealt from a deck of 52 cards?

1. Answer:  possible hands

How many different 5 card hands can be dealt from a deck of 52 cards if all five of these cards are clubs?

1. Answer:  possible hands

How many different 5 card hands can be dealt from a deck of 52 cards if the hand consists of exactly 2 aces?

1. Answer:  possible hands

How many different 5 card hands can be dealt from a deck of 52 cards if the hand consists of exactly 2 aces and exactly 2 kings?

1. Answer:  possible hands

Answer 1

Here are some suggestions for number of 5-card hands containing exactly two aces.

There are exactly 4 aces in the deck. How many ways can you choose two of them?

This is represented as ₄C₂ (read as four choose two, or Combination of 4 things taken two at a time).

That value is 4(3)/2! = 6.

The rest of the hand contains 3 cards that are not aces. That means the rest of the hand will be filled out by taking 3 cards from the remaining 48 (subtracted the 4 aces from 52 cards in a deck). The number of ways to do that is ₄₈C₃. To calculate that value manually 48(47)(46) / 3!

Now you have to pair up those events using the fundamental counting principle. If there are m ways to do the 1st event and n ways to do the 2nd event then the number of possibilities is m times n.

So the number of 5-card hands dealt from a deck of 52 cards that contains exactly 2 aces is:

₄C_{2 48}C₃

Here is a Desmos graph showing the calculation.

desmos.com/calculator/yeqeiyt8xm

Answer 2

All possible hands: 52! / (5!) / [ (52 - 5)! ]

All possible club hands: 13! / (5!) / [ (13 - 5)! ]

All possible hands with one pair of aces: {50! / (3!) / [ (50 - 3)! ] } * (4*3/2)

All possible hands with one pair of aces and one pair of kings: 48 * (4*3/2) * (4*3/2) = 48 * 6 * 6 = 1,728

Answer 3

I will answer the first question...you need to answer the others for yourself and if you have specific questions, make a comment to that effect.

This is a straight application of the binomial coefficient which in general is the number of subgroups of size k which can be chosen from a set of n different things. In this case ₅₂C₅=52!/(5!*47!)

The other questions are modifications of the approach which is called the hypergeometric distribution which splits the numerator into 2 distinct groups along with the notion that the number of ways of doing 2 consecutive (independent) things is the product of the ways of doing each individual thing.