- SLIDES 2 to 6 are a review the basics of combinations and permutations
- SLIDES 7 to 16 deal with poker combinations

SLIDE 2

In how many ways can 8 books be arranged on a shelf if 3 particular books must be together?

There are 8 books. We grouped 3 of the books together. (If you imagine that the 3 books are together in a bag, this might be helpful to you.) So now we have 6 objects. We can shuffle the 6 objects 6! ways for a different arrangement. We can also shuffle the 3 books in the bag 3! ways for a different arrangement.

Answer: 6!3! = 4320 ways

SLIDE 3-4

There are 10 football teams in a certain conference. How many games must be played if each team is to play every other team just once?

SLIDE 3 is wrong. SLIDE 4 is correct.

To understand this problem, let's simplify the problem to 3 football teams. Let's imagine Mr.K's team, Rence's team, and AnhThi's team are together in a conference, and they're to play each other in a game just once. Mr.K versus Rence in one game, and Mr.K versus AnhThi in another game. So far, there are 2 games played. Rence versus AnhThi in a game, so the maximum number of games that can be played between the 3 teams is 2+1 = 3. Here we see a pattern that we're adding. WE'RE NOT USING FACTORIAL. The student who did SLIDE 3 made that mistake and got the wrong answer.

So if there are 10 teams that are to play against each other and they're to play each other in a game just once, then the first team can play any of the 9 other teams, the second team can play any of the 8 other teams, the third team can play any of the 7 other teams, the fourth team can play any of the 6 other team, etc. And so, the maximum possible number of total games played in a game is:

Answer: 9+8+7+6+5+4+3+2+1 = 45.

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Another way to solve the problem is to see that there are 10 games and 2 teams are chosen to play against each other. This can be expressed as:

Answer: 10C2 = 10!/(8!2!) = 45

SLIDE 5

There are 9 chairs in a row. In how many ways can 4 students be seated in consecutive chairs? (Hint: First find the number of ways of choosing 4 consecutive chairs.)

First, we counted the number of ways of choosing 4 consecutive chairs. By moving the red "container" (that holds only 4 chairs) at a time to the right, we find that there are 6 ways to get 4 consecutive chairs.

Now, we ask ourselves, how many ways can 4 students sit in a row of 4 chairs? Answer is 4!.

6*4! = 144

SLIDE 6

Seven people reach a fork in a road. In how many ways can they continue their walk so that 4 go one way and 3 the other?

There are 7 people. These are your slots.

_ _ _ _ _ _ _

Each person (represented by a slot) either goes left or right because there's a fork in the road, but there must be 4 that go in one direction, while the other 3 go the other direction.

_ _ _ _ _ _ _

L L L L R R R

Remember, the formula for permutations of non-distinguishable objects says:

n! / (k1! k2! k3!)

n is the number of objects that contain k1, k2, k3..., which are non-distinguishable objects.

Answer: 7!/(4!3!) = 35

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Another way of solving the problem is to choose 4 people from the 7 to go in one direction, while you choose 3 people to go in the other direction, which is expressed like so:

7C4 * 3C3 = 35

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The rest of today's lesson is the "juice" of today's lesson.

POKER COMBINATIONS

Given a standard deck of 52 cards, how many ways are there to draw 5 cards to obtain each hand? (SLIDE 7)

In other words, "how many different 5 card poker hands are possible?"

Well, there are 52 cards and we're choosing 5 from them. This is expressed as:

52C5 = 2 598 960 ways.

(a) Royal Flush [ace, king, queen, jack, ten in the same suit]

There is only one way to get the sequence: ace, king, queen, jack, and ten if they're all in the same suit. This is expressed as:

4C1 = 4.

(b) Straight Flush [five cards in sequence and of the same suit, but not ace, king, queen, jack, ten] (SLIDE 8)

* This problem is similar to the problem with the red container and chairs.

* The black work on SLIDE 8 is wrong, by the way.

Here are your "slots."

A 2 3 4 5 6 7 8 9 10 J Q K

Because of my limitations of only describing things to you and not by means of animation, please bear with me.

Imagine the container again and this time the container can hold up to five slots. Starting from [A, 2, 3, 4, 5], we move our container to the right and count how many ways the container can hold consecutive slots. We count that there are 9 ways to have five different slots consecutively in the container.

By definition, a 'straight flush' must be of the same suit, so we can only use one out of the four suits. This can be expressed as 4C1, "4 choose 1". (From the four suits, only one of them is being chosen.)

Answer: 9 * 4C1 = 36

(c) Four of a kind [four cards of one face value and one other card] (SLIDE 9)

The first four cards...

There are 13 different face values per suit, and 1 face value is being chosen of that thirteen; this is expressed as 13C1. In total, there are 4 face values of each suit, and any of those 4 face values of each suit can be chosen; this is expressed as 4C4.

The fifth card...

The fifth card can't have the same face value of the first four cards, which leaves us with 12 different face values remaining. Of those 12 face values, we just want 1 of them; this is expressed as 12C1. There are 4 suits, only 1 will be chosen; this is expressed as 4C1.

Answer: 13C1 * 4C4 * 12C1*4C1 = 48

(d) Full house [3 cards of one face value and 2 cards of another face value] (SLIDE 10)

"3 cards of one face value"

There are 13 different face values, and only 1 of the face values is being chosen; this is expressed as 13C1. There are 4 different suites, and 3 of one face value is being chosen; this is expressed as 4C3. Now we have our three cards of one face value.

"2 cards of another face value"

Since, by definition, a full house includes 3 cards of one face value plus 2 cards of ANOTHER face value, we can only choose from the remaining 12 face values that hasn't yet been chosen by the first 3 cards. Of the remaining 12 face values, only 1 is being chosen. There are 4 different suits, and only 2 of one face value are being chosen; this is expressed as 4C2. Now we have our 2 cards of another face value.

Answer: 13P2 * 4C3 * 4C2 = 3744.

HOMEWORK

- You should be at Exercise 35.
- Slides 11 - 16.

NEXT SCRIBE

By the process of elimination, Eleven is scribe.

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