### scribe

hey everyone this is jessica i just found out today that i was scribe and i don't see my name anywhere. but thanks for the heads up khan. so today we had two math periods and for those who didn't go to class today, which shouldn't be too many people cause i think i saw everyone, Mr.K gave us notes to write in our dictionaries. After we were given the notes Mr. K gave us a stencil to work on in groups. I got to say the stencil is pretty difficult so if you missed class better get some help from Mr. K.

These are the notes that were given to us

Combinations:An arrangement of objects where order doesn't matter.

Notation: nCr or (n over r) [sometimes called the "choose" formula]

read as: "n choose r"

formula nCr = n! / (n-r)!r!

n is the number of objects to choose from.r is the number of objects being arranged.

meaning: from a set of n objects, how many different ways can you choose r at a time if the order they are chosen in doesn't matter.

example: how many different lotto 6 / 49 tickets can be sold?

Solution 1: As a permutation of non-distinguishable objects we imagine ourselves saying "yes" to 6 of the numbers from 1 to 49 and "no" to the rest.

ie. we say "no" (n) to 49 - 6 or 43 the number of ways to do this is the same as the number of different ways to spell a 49 letter word made up of 6 y's and 43 n's.

49! / 6!43! = 13 983 816

Solution 2: as a combination

49C6 = 49! / (49 - 6)!6! = 49! / 43!6! = 13 983 816

those were the only notes that were given to us and for the rest of the morning class and afternoon class we were given time to work on our stencils.

there was a question on the stencil that a number of us were asking Mr. K about so he explained to us how to solve it. It was question number 4 and here it is.

4.(a) How many numbers of 5 different digits each can be formed from the digits 0, 1, 2, 3, 4, 5, 6?

(b) How many of these numbers are even?

(c) How many of these numbers are divisable by 5?

(a) answer is 6 * 6 * 5 * 4 * 3 = 2160

the reason that the first number of the 5 digits has only 6 choices is because you can't include 0 to be the first number. The second digit has 6 choices because we can include the 0 back into the number of choices and there would only be 6 choices left, then 5, 4 and 3.

(b) answer is 360 + 900 = 1260

In this question the 0 is a problem again because it is also an even number so instead we take it out. Now we only have 3 choices for the last digit to be even and since the first digit also can't be 0 the number of choices we have is 5. The second digit is also a5 because we can include the 0 back into the choices the 4, and 3. 5 * 5 * 4 * 3* 3 = 900

We do this again but instead of the last choices being 3 it is 1 because we put 0 as the last digit. then the first digit has 6 choices then 5, 4 and 3. 6 * 5 *4 * 3 * 1 = 360

Now all that is left to do is add those numbers up to get the total number of combinations with the last digit being even.

(c) answer is 360 + 300 = 660

In this last question we are asked how many numbers would be divisable by 5, now we can all identify when a number is divisable by 5 when the last digit is a 5 or 0. Like in the other questions 0 becomes a problem, if the last digit is 0 the number of choices for the first digit is 6 then 5, 4, and 3. 6 * 5 * 4 * 3 * 1 = 360

When the last number we choose is 5 we have to take out the 0 for the first digit and the number of choices becomes 5 and the second digit is also 5 because we know that we can put 0 back into the number of choices then 4 and 3. 5 * 5 * 4 * 3 * 1 = 300

The last thing to do is add those two numbers up to get the number of combinations that will be divisable by 5.

Now that im done I hope that sorta helped anyone who didn't understand what we were doing in class today and those who missed it. I tried the best I could to make this understandable, I actually had to do it twice because my first one got deleted. well thats all. tomorrow's scribe will be muuxi.

These are the notes that were given to us

Combinations:An arrangement of objects where order doesn't matter.

Notation: nCr or (n over r) [sometimes called the "choose" formula]

read as: "n choose r"

formula nCr = n! / (n-r)!r!

n is the number of objects to choose from.r is the number of objects being arranged.

meaning: from a set of n objects, how many different ways can you choose r at a time if the order they are chosen in doesn't matter.

example: how many different lotto 6 / 49 tickets can be sold?

Solution 1: As a permutation of non-distinguishable objects we imagine ourselves saying "yes" to 6 of the numbers from 1 to 49 and "no" to the rest.

ie. we say "no" (n) to 49 - 6 or 43 the number of ways to do this is the same as the number of different ways to spell a 49 letter word made up of 6 y's and 43 n's.

49! / 6!43! = 13 983 816

Solution 2: as a combination

49C6 = 49! / (49 - 6)!6! = 49! / 43!6! = 13 983 816

those were the only notes that were given to us and for the rest of the morning class and afternoon class we were given time to work on our stencils.

there was a question on the stencil that a number of us were asking Mr. K about so he explained to us how to solve it. It was question number 4 and here it is.

4.(a) How many numbers of 5 different digits each can be formed from the digits 0, 1, 2, 3, 4, 5, 6?

(b) How many of these numbers are even?

(c) How many of these numbers are divisable by 5?

(a) answer is 6 * 6 * 5 * 4 * 3 = 2160

the reason that the first number of the 5 digits has only 6 choices is because you can't include 0 to be the first number. The second digit has 6 choices because we can include the 0 back into the number of choices and there would only be 6 choices left, then 5, 4 and 3.

(b) answer is 360 + 900 = 1260

In this question the 0 is a problem again because it is also an even number so instead we take it out. Now we only have 3 choices for the last digit to be even and since the first digit also can't be 0 the number of choices we have is 5. The second digit is also a5 because we can include the 0 back into the choices the 4, and 3. 5 * 5 * 4 * 3* 3 = 900

We do this again but instead of the last choices being 3 it is 1 because we put 0 as the last digit. then the first digit has 6 choices then 5, 4 and 3. 6 * 5 *4 * 3 * 1 = 360

Now all that is left to do is add those numbers up to get the total number of combinations with the last digit being even.

(c) answer is 360 + 300 = 660

In this last question we are asked how many numbers would be divisable by 5, now we can all identify when a number is divisable by 5 when the last digit is a 5 or 0. Like in the other questions 0 becomes a problem, if the last digit is 0 the number of choices for the first digit is 6 then 5, 4, and 3. 6 * 5 * 4 * 3 * 1 = 360

When the last number we choose is 5 we have to take out the 0 for the first digit and the number of choices becomes 5 and the second digit is also 5 because we know that we can put 0 back into the number of choices then 4 and 3. 5 * 5 * 4 * 3 * 1 = 300

The last thing to do is add those two numbers up to get the number of combinations that will be divisable by 5.

Now that im done I hope that sorta helped anyone who didn't understand what we were doing in class today and those who missed it. I tried the best I could to make this understandable, I actually had to do it twice because my first one got deleted. well thats all. tomorrow's scribe will be muuxi.

Good scribe post Jessica!

I really liked the way you took one of the more difficult questions and explained how to do it step-by-step. This will be a great resource for people studying for the test and exam.

Way to go!

Posted by Mr. Kuropatwa | 3/01/2006 10:16 PM

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