Poker Hands Permutations Combinations
- Poker Hands Permutations Combinations Chart
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- Poker Hands Permutations Combinations Worksheet
- Poker Hands Permutations Combinations Calculator
Summary
Students learn how to use permutations and combinations and when it is appropriate to use each.
Combinations and Permutations How many different poker hands are there? How many different pizza orders can be made? How many different ways can this work schedule be filled out? How many different ways are there to arrange your books in a bookshelf? These are all examples of combinations and permutations. And knowing how to calculate them is a helpful tool for decision making. The bonus spins are Permutations And Combinations Poker Hands subject to wagering requirements of 30 x bonus amount. The maximum bet allowed when using bonus money is of €5 per spin or €0.50 per bet line, until the wagering requirements have been fully fulfilled. This bonus offer is subject to country restrictions. The order in most lottery draws does not matter. If we examine the poker example further: a poker hand can be described as a 5-combination of cards from a 52-card deck. The 5 cards of the hand are all distinct, and the order of cards in the hand does not matter so it is a combinatorial problem.
Background for Teachers
I will just need a pack of cards, a handful of dice (with various numbers of sides) and a whiteboard. Any examples I use will be listed below.
Student Prior Knowledge
Students should have some familiarity with the factorial.
Intended Learning Outcomes
Students will understand the fundamental differences between permutations and combinations and be able to use this knowledge to determine when a permutation is to be used and when a combination is to be used.
Instructional Procedures
I will begin with the dice, looking to teach the fundamental counting principle. I will use a d6, d8, and d20, rolling the smallest die first and ask my students how many combinations of numbers I can get rolling these dice. In rolling the first dice, how many possible numbers can I get? Rolling the second, and then the third? Give students time to work out a solution. Once they have an idea worked up, state the fundamental counting principle: If 2 events occur in order and the first can occur in m ways, while the second can occur in n ways (after the first has occurred) the two events can occur in order in m*n ways. Proceed through more examples, asking specific students. When they respond, I will ask them how they calculated that answer. Let's say I'm working with a drum machine that contains 11 different bass drum samples. On the bass drum channel, I can apply 1 of 4 different effects (distortion, reverb, delay, chorus), how many different bass drum sounds can I get out of this drum machine? Now that I'm done working on my drum machine, I hear the ice cream man coming. He's got 3 different cones and 5 different flavors, how many different ice cream treats can I get? After all of that ice cream, I'm parched. I head to the Kwik-e-mart where they have 21 different kinds of soda on tap with 6 different add-in flavors (cherry, vanilla, etc.), how many different kinds of drink can I get? Do we count no add-in as an option? Once the students seem comfortable with the fundamental counting principle, we can begin discussing permutations. I tell the students that I have just 10 songs on my iPod (a blatant lie, but I don't want them to deal with numbers that are too big). If I put my iPod on shuffle so it will randomly pick a which song plays, and it won't play a song again until it has gone through all of them, I ask my students how many possible sequences of songs are there if I listen until all 10 are played? I don't think 'a lot' is an acceptable answer. If they don't seem to be moving towards an answer, I'll remind them of the principle we just learned. How many possible first songs are there? (psst, 10) Having played the first song, how many possible second songs are there? And so on. There is a connection here between this concept of permutation and the counting principle we've learned. The first song is the first event that can occur, the second song is the second event and on it goes, so we just keep multiplying 10*9*8*...*3*2*1. I will ask if this loooks familiar to anyone? Does it, possibly, make you want to EXCLAIM, 'I know what that is!!!' It's a factorial. Factorials are what we use to calculate permutations. If I have n objects, I can have n! permutations of those objects. That's the key part to remember. If I have 10 songs, I can have 10! (3,628,800) possible sequences of songs while listening on shuffle--seems like a lot for just 10 songs. Now, let's say I want to make a playlist of 3 of those 10 songs. It could be any 3 distinct songs, but the order of the playlist does matter--that is if I have 2 playlists with the same 3 songs, but they are in a different order in each playlist, then those are different playlists. If I just had 3 songs to choose from altogether, how many different playlists of 3 could I make? How many different songs could be the first song, and once that is set how many could be the second, etc. I could end up with 3! (or 6) different playlists. But I've got more than 3 songs to choose from, so if I've got 10 songs on my ipod, how many different songs could be the first and how many could be the second? How many distinct 3-song playlists could I come up with from my massive library of 10 songs? I would have 10*9*8 = 720. So how might we calculate this? How could we come up with a formula that will give me a permutation of r objects selected from n objects. The following section consists of a number of questions designed to guide students to the formula for 'n choose r permutations.' Sufficient wait time should be used after each question to see if the students can come up with the formula on their own. In the case of my playlist, I wanted to make a 3-song playlist (r = 3) from 10 songs (n = 10), and we ended up multiplying the last 3 terms of the factorial together. How could we express that in a general notation, in terms of n and r? How about reduction of fractions. If I wanted to calculate this 'choose 3 from 10' permutation, I might write it as (1*2*3*4*5*6*7*8*9*10)/(1*2*3*4*5*6*7), or more concisely, 10!/7!. Can we relate these back to n and r? The general formula is n!/(n - r)! because by dividing by (n -- r)! we are reducing it to just the last r terms of n! This is how we calculate 'from n objects, choose r.' It is usually written as P(n, r). Now what about dealing cards? I ask, if I deal a poker hand (5 cards) from a deck of 52 cards, how many permutations would I have? Wait for response, students should answer 52!/47! = 311,875,200. That's how many permutations of cards could be dealt. Remind the students, though, that it doesn't matter what order the cards are dealt in. A full house is still a full house, no matter how the cards were dealt. I ask students how many permutations (just in factorial notation) would there be of the 5 cards that are dealt--just looking at those 5 cards alone? It would be 5! permutations, but each of those 5! permutations is just a different arrangement of a single combination. A combination is like permutation, but order doesn't matter. A hand consisting of the ace of spades, queen of hearts, 2 of clubs, jack of diamonds, and 7 of hearts is exactly the same as a hand consisting of the queen of hearts, jack of diamonds, ace of spades, 7 of hearts, and 2 of clubs--and neither will get you anywhere in poker. There are significantly fewer combinations than there are permutations. In fact, in this example, for every 1 combination, there are 5! permutations. Thus, if we were to find a general formula for picking combinations of r objects from a set of n objects, we would come up with P(n, r)/r!, or more formally, C(n, r) = n!/(r!*(n -- r)!). Notice that C(n, r) is how we write combination of r from n. I will then ask my students, what is the important distinction between permutations and combinations? Their answer should be somewhere along the lines of whether order matters or not. Students will pair up into squish groups and then work the following examples. Each group will raise their hands once they've got an answer and I will call on someone once all groups have their hands up. To answer, students need to identify whether a permutation should be used, or a combination, evaluate the answer, and describe the steps they took to arrive at that answer. In how many different ways can 6 people be seated in a row of 6 chairs? (6! = 720, permutation) In how many ways can 3 pizza toppings be chosen from a selection of 12? (12!/(9!*3!) = 220, combination) If 100 people sign up to be beta testers for a new program, and only 4 are chosen, how many different groups of 4 could there be to beta test the program? (100!/(96!*4!) = 3,921,225, combination) In how many ways can a president, vice president, secretary, and treasurer be chosen from a class of 30? (30!/26! = 657,720, permutation) Once this exercise is done, assign homework and let the students be on their way.
Poker Hands Permutations Combinations Chart
Strategies for Diverse Learners
Poker Hands Permutations Combinations List
By drawing upon a wide variety of examples, I hope to engage all students by appealing to personal interests. When students get to see this kind of mathematics in action, they are more apt to learn it.
Assessment Plan
tudents will assess themselves and their squish partners during the example exercises at the end of the class. I will also be wary of student responses during this time and try to catch any misconceptions or groups that don't have the right answers. A bell quiz the following class day will consist of the question: How many 5-card hands can be dealt from a deck of 52 cards? We will also cover the homework the next day, allowing students to self-evaluate their work.