Saturday, April 21, 2007


HI guys! I'm Sandy and I'm once again scribe for today. I just wanted to say THANK YOU so much John for volunteering to be scribe for Thursday when I needed someone. I'd also like to apologize for the scribe post being up late. It's been a rough weekend, so again, my apologies.


At the beginning of class we talked about different things we could do for our DEV Projects. To make fancy fractions like Mr. K makes on Apple Works, you can either download MathType or you can use Microsoft Word. Personally, I use Microsoft Word 2007, but I'm not sure if the older versions can do that as well. To upload photos you can use Flickr, most of you know about Photobucket, there's also Fileden, Freewebtown and Prizaar.

Then we started with our lesson!

Follow the butterflies!
You need to make a four letter word. So basically there's 7 letters you can use all together. The first letter you have 7 options to pick from, then one is being used so for the next spot you only have 6 options. Then the next is 5 options and then 4 options.
The red answer is the correct answer. At the beginning there has to be a vowel and at the end there has to be a consonant. Regardless of which letters those are, there a two letters that have to be in each spot. Which means, there's only 5 letters that can go in the second spot and 4 in the other since two are already being used.
Out of the 7 letters, there are 3 consonants and 4 vowels. To make them alternate they either have to be vowel, consonant, vowel, consonant or consonant, vowel, consonant, vowel. So, for the first, once 1 vowel is used the third letter only has 3 options left. Same goes for the consonants, once one consonant is used, there's only 2 options left. Once you've figured it out, you add the two together and it gives you your answer.

When it comes to circles things change a bit. To break it down, the order in which people or objects are placed around a table is the only thing that matters. Rotating them doesn't make a different, frankly they're just in a different position around the table with the exact SAME order. The same person will be on their right and the same person will be on their left. So for these kind of questions the formula to find it is (n-1)! (Where "!" means Factorial not "I really, really mean it"). On this slide the circles in RED are the same and the circles in BLUE are the same.
HOWEVER... tables are different from bracelets. With bracelets, you're able to turn them over. So technically each arrangement actually represents 2 of them. The same applies for necklaces and other things that can be flipped. In these special cases, the formula you use is (n-1)!/2.

With this questions, you can pretty much treat the married couples as ONE person since they always have to be across from each other. Once one sits, the other one has no choice where to sit. So since there are 4 couples, each treated as ONE the answer to this question is (4-1)! --> 3!. Or you can draw it like above.
Since the boys are Gentlemen and we, are the ladies we'll be seated first. So the first lady seated is the reference point and she determines how everyone else can be seated. Therefore they'll be (4-1)!. Now you have four spots left for the men to sit. Since the reference point has already been placed the first spot can be used by 4 men, then the next only has 3 men to sit there and so on and so forth. So the men will be 4!. Multiplying those together you'll arrive at 144.


In the afternoon we were lucky to accompanied by Mr. K children Kas and Emilia! (Who by the way, are soo adorable). Anyway, Emilia sat with me and Kas sat with Aichelle. (: Moving along!

Instead of it being 6!, it's (6)(6)(5)(4)(3)(2)(1). This is because if 0(zero) was the first number, it wouldn't be 5 digit numbers anymore, it would only be a 4 digit number. So for the first number, there's only 6 numbers out of the 7 to use. For the second number, no matter what number the first is, there's also 6 options because the zero can be used there now. Then it decreases from there.
The second question is actually not the correct one. This is:

Basically to make it easy you do one set for 0(zero) and another for the other numbers. For 0, no matter what zero has to be the last number which gives you only 1 option. Then for the first number you have 6 other options. Then so on and so forth. For the second part, you have 3 options but no matter which you pick, you only use 1 of them. The first number HAS to be 5 oppose to 6 like the previous because 0 cannot be the first digit otherwise it won't be a 5 digit number anymore. Then you add the two together. (:

Again, that's the wrong answer and the correct one is this:

It's pretty much like the previous part to this question, but instead of 2, 4 and 6, we use 5, based on what the question asks.

This question originally said, How many necklaces can be made from 12 beads of different colours? Thanks to Jojo and his brilliant mind, we realized the actual answer was 1. With 12 beads of different colour, we can make one necklace. It doesn't specify how many different colours. Then Mr. K changed it! But of course, Jojo gets the credit for recognizing it.
To solve this question is fairly straight forward. You have 12 beads, but it's a circle so the first bead is the reference point. So it's (12-1)! Also, since it's a neckalce, if it's flipped it's the same. Therefore you divide by 2 and there's your answer.

This is a Combination type of question. These are when you arrange objects and don't care what order they're in. However, afterwards you DO care what order they're in. So first you just find the different ways that a necklace of 12 beads can be made with 18 beads. (18!/12!6!) Then, for each of those different ways, you need to find out how many different ways the beads can be arranged. So you multiply it by (12-1)!/2. You simplify and come up with your answers.

Since the students have to be right beside each other everytime (consecutive), there's only 6 ways that they can be arranged in the 8 seats (Shown by Aichelle at the bottom of slide). Since there are 4 students, the first seat has 4 students to sit there, then the next 3, the next 2, and the next only 1. So it's 4! multiplied by the 6 different ways they can sit in the 8 seats.

The 3 books that always have to be together, you should just put into a bag until later on in the question (Mr. K's blue bag). Considering the 3 books in the blue bag as 1, the different ways they can be positioned is 6!. Then, if you take the books out, the different ways the 3 books can be arranged is 3!. Multiplying them together you get the answer. (:

Hopefully I explained everything enough for people to understand. I again apologize for the late post. Usually if I have the weekend I like to post it up on that friday. But since things got in the way, it never happened. So again, sorry.
AND.. the next scribe will be the super awesome girl that sits next to me in class, Mel!

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