We had a fairly long chat about Justus' awesome scribe post, and his BOB before that and how awesome that was, and Mr. K introduced for the first time to us, a "Scribing Hall of Fame" and how to get inducted into this Hall of Fame. For you to be accepted, you must have a vote with the class, and if enough students vote to induct your special scribe post then congratulations you're scribe post is now a Hall o' Famer. Justus, you have my vote, thumbs way up. If you want to see what it takes to be just like Justus, then just read all the past scribe posts that are in the Hall of Fame. Found just a click away at this link: http://thescribepost.pbwiki.com/HallOfFame

At the start of today's class Mr. K had us confused on why combination locks shouldn't be called "combination" and how they should be called permutation locks. This had me quite confused. I started to wonder, that if three different numbers in different orders isn't a combination, then what is? If you want to find out, you must read on. At the start of class, we were given questions to do, these are found on slide 2 of today's slide show. These questions were similar to last class' questions, refer Justus' scribe post at: http://pc40sw08.blogspot.com/2008/04/intro-to-combinatorics.html

There was one difference though, these new questions had the number zero included as an option. With the question: "How many numbers of 5 different digits each can be formed from the digits 0, 1, 2, 3, 4, 5, 6?" Using a "slot system" (as found in the 2nd slide) is easiest to use. The first slot can usually be any number, but not if that number is zero, and this is where the confusion and difficulty kicks in. It's because if we are referring to a 3 digit number, we can't say that number is 012, because this is just a 2 digit number with a zero placed at the front, so the zero has no real value. If you included zero as a first digit option, then it would be a 4 digit number, not 5 which was asked for in the question. The first digit therefore only has 6 options. Now for the 2nd slot, you can use zero, because it will now have value (ex. 01 = one, but 10 = ten), so you have 6 options, instead of 7 because one option was used for the first slot. The 3rd slot would have one less options, which would be 5 options, and so on until all slots have been taken up. Multiply all these number and you should get 2160, which is the amount of options you have of 5 digit numbers created from the number 0, 1, 2, 3, 4, 5, and 6.

For part b of the question we had to find how many of the 2160 options are even numbers, which means the number would have to end in 0, 2, 4, or 6. This question has 2 parts to it, which means 2 different slot systems, as found in slide 2. One slot system would be numbers that don't end in zero, if these numbers don't end in zero, you would have 2 less options for the first digit. (options: 1, 2, 3, 4, 5, 6. (total 6 options)) and it wouldn't start with the number used in the last digit (total: 5 options), it would be easier to start with the last digit on this question. The slots would then be 5, 5, 4, 3, 3. The last digit would be 3 instead of 4 because you're not ending it in zero this time (leave the 2nd slot system for numbers ending in zero). When the numbers in this slot system are multiplied this would give you 900 options. For the 2nd option you would use numbers ending in only zero, so the last slot would be a 1 (the only option being zero). Since the last slot is zero, the first slot can have 6 options (1, 2, 3, 4, 5, 6) because the last slot doesn't take up any of these numbers. The next slot would be 5, then the 3rd slot would be 4, and so on, with the last slot being 1. When all slots are multiplied, you're left with 360 options. These options are even numbers ending with zero. Now add the total number of options from 1st slot system, and 2nd slot system, and you would have 1260 options which would be even numbers.

Part C is "How many of these numbers are divisible by 5?" For a number to be divisible by 5 it would have to end in either 0, or 5. We now know that if it has to end in zero, we need to do 2 different slot systems, a system ending in 0, and another system, not ending in 0, for this particular question, the only other number is 5. The first system will end in zero: so our first slot would be 6 different options, then 2nd slot would be 4 options because zero is used and 5 is used for the different slot system, and another options is used for the 1st slot of this system. The 2rd slot would be one less, and so on, with the last slot being 1 option. When all slot are multiplied together, you get 360 options. Now for the next slot system, which ends in 1 options, which is 5. This would give us 5 options for slot 1, because it can't include 0 or 5. Then this would give us 5 options for slot 2, because it cant include, 5, or the digit used for slot 1, then 3rd slot would be 4, and so on, with the last slot having 1 options. When multiplied all together we have 300 different options. Now add the results from both slot systems and we have 660 options that are divisible by 5.

In this way, these questions are a problem, because they include the number zero. If these systems end in 0, there's not much trouble, but if they don't be positive that the leading number won't be zero.

We then worked on factorials. On slide 3. We would simplify these by expanding them, then reducing. (Ex. 8!/7! = 8 because 8!/7! = (8 x 7!)/7! and the 7!'s reduce to get 8) This is quite simple, and if you don't understand factorials, refer to Justus' scribe post, the link is mentioned in the 2nd paragraph.

In the factorial notation: n! The variable n, can't be negative or a decimal, because of a domain error. Remember that the factorial notation is a definition.

We did another problem similar to the first on slide 4. All you have to do it pay attention to what the question is asking you, and it's quite simple.

Permutation is an ordered arrangement of objects without repetition, as found on slide 5. Although permutations is in combination so permutation is an ordered arrangement of objects without repetition, in combination.

We also learned an equation entitled the "Pick" formula, which is like the slot system but simplified, found on slide 5. Mr. K says we won't be using this equation a lot. I take this as to not sweat over this equation.

In a combination of numbers, the order doesn't matter and it can repeat numbers, but in a permutation order does matter, and there is no repetition. This is pretty much why combination locks should be called permutation locks instead.

This is everything that has to do about anything in our class today. Enjoy and don't stress. Sorry if this blog seemed a bit rushed, I just had to rush it so I can get to work on time. Apologies.

One last thing, the next scribe will be Paul.

-Francis

## Monday, April 28, 2008

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