Wednesday, May 13, 2009

Bob = 8 Permutations and 5 Combinations

It's Aldrin S. here.

I found this unit to be quite interesting, yet it is the first I encountered that is actually difficult to master. You really need to use your head for this unit. There are myriads of word problems that can be made, which do not allow for a way to be answered in a consistent manner. That is the beauty of this unit. One must really understand patterns and know what to do based on the information that is provided by the question. It is very important to KNOW what the question is asking for since this is usually why students get wrong answers. They misinterpret what is being asked, which leads them to doing their math wrong.

Personally, I find this unit hard. I am pretty rusty and can easily get owned by a word problem. My main problem deals with the poker combinations. The poker combination questions are quite killer. I think I just do not know how to begin solving these questions. I looked at wikipedia and analyzed the combinations of poker hands. The mathematical expressions of absolute frequencies that are stated are very understandable, except when those values are not present within my sight, I get owned basdly. As for everything else, I think I’m good. I’m just 75% prepared for the test.

What is a permutation?
The number of ways things can be arranged where order is a factor.

What is a combination?
The number of ways things can be chosen where order does not matter. Basically, it is a selection. I want ABC. No wait, I want CAB, oh wait.. They're the same thing.

What is the difference between a permutation and combination?
Order matters in a permutation whereas it doesn't in a combination. A combination is just a selection of things or people.

Do you explicitly label your numeric values when showing your work? (the thing Mr. K warns us that we can lose marks for)
4 x 2 x 1 = 8
See that 4 over there, the 4 represents the number of ways ... Something like that lol.

Do you know to divide the product by two for a bracelet when you use the circular permutation formula for finding the number of arrangement for beads?
The wording of that question is not that great, but remember use the special circular formula where u subtract one from the original number, then find the factorial of that, then divide by two.

Do you know lots about the binomial theorem and Pascal's Triangle? cuz i don't jk
Know how to find the terms and what term consists of x^7 for example. Know the patterns and what not too!

Are you memorizing how to do things for word problems?
because that's a no no.

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Poker Combinations

Given a standard deck of 52 cards, how many ways are there to draw 5 cards to obtain each hand.

There are 52 cards. You must choose 5. Order does not matter, nor does the type of hand.

a) Royal Flush

There are 4 suits and you can only get a Royal Flush once in each suit.

b) Straight Flush

There are 4 suits. How many ways can you get a straight in one suit? You can do it in 10 ways but Mr. K eliminated the Royal Flush, so it's just 9 now for each suit. 9 ways x 4 suits = 36. Or you can do this. 10C1 * 4C1 - 4C1 because the 10C1 * 4C1 gets all the ways to make straight flushes. Then subtracting 4C1 is actually subtracting the number of ways to get a royal flush.

c) Four of a Kind

There are four suits. How many cards are in each suit? How many cards are left after 4 are used?

d) Full House

You choose one card from one suit, which has 13 cards. In order to obtain a TRIPLE, you should know that there are 4 suits and from 4, you choose 3. It should be like 13C1 * 4C3 . You see, these guys are like brothers. The 13 is what it is because it corresponds to each suit. Disregard multiplying 13 by 4. Now you just choose 1 card from the 13. So for 4C3, this is the DUDE that makes the 13C1 affected by all suits. They're like brothers you see. There are 4 suits and 3 must be chosen.

As for the DOUBLE, you should know that there are no longer 13 numbers from each suit to choose from since the TRIPLE eliminated those choices. *There are 12 now* So there are 4 suits, and you choose 2. It should be like 12C1 * 4C2. You see, because of the TRIPLES, it eliminated a number to make doubles ! So its like a have a triple of 5s. Now I no longer have 5s to make doubles and that affects each suit. Aww too bad. So as for 4C2, it makes the 12C1 you know, be affected by all 4 suits.

e) Flush - toilet flush hehe

In a suit, there are 13 cards. You must choose 5 from one suit. There are also 4 suits. It should be like 13C5 * 4C1. Remember the 4C# makes the 13C# or 12C# or whatever it is correspond to all the suits ! Uhmm. Or, you can say the 4C# makes what you choose be available from 4 suits. If it was 3C#, then I can only choose from 3 suits.
Now you just subtract the number of Straight flushes ! It says ignore Straight flushes and Royal flushes. We can just ignore the royal flushes since the straight flushes takes care of the royal flushes.
So for the straight flushes, figure out the number of ways a straight can be achieved, then multiply that value by the number of suits. There are 10 ways in each suit since 1-5, 2-6, 3-7, 4-8, 5-9, 6-10, 7-11, 8-12, 9-13, 10-1. It should be like 10 * 4C1

f) Straight

They don't have to be in the same suit and he doesn't want Straight Flushes. So basically, it's like 10 * 4C1 * 4C1 * 4C1 * 4C1 * 4C1 - 10 * 4C1. So there are 10 ways to make a straight. Multiply that by 4C1 ^ 5 because each 4C1 is a single event in which you pick a successive card from a choice of 4 suits! Basically, it's like this. I choose ace of hearts, two of spades, three of spades, four of clubs, 5 of hearts. There was no diamonds because there doesn't have to be diamonds. The spades was repeated because it can repeat. You always have 4 suits to choose from when choosing 1 card. You have to do this 5 times because there are 5 cards ! As for the subtraction, it's evident from the question earlier.

*Imma just put very short explanations now since you should be good at this by now*

g) Three Of A Kind - One of a Kind times 3 :P

13C1 * 4C3 * 12C2 * 4C1 * 4C1

13 cards per suit, I choose 3 cards from the 4 suits. This leaves me with 12C2 because if it was 13, then I'd still be able to get the card I got earlier. However, That would make me get Four of a kind instead of Three of a kind. So I choose two cards while the 4C1^2 are the two events where I choose from 4 suits. So thats 2 times I choose !

h) Two Pairs

13C2 * 4C2 * 4C2* 11C1 * 4C1
Boo yah ! No explanation ! You're a master now !

Forgot to do One Pair and No pair. You guys can do it without my help now cuz you're pro now !

Check if your delicious link is here !


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