## Monday, April 30, 2007

### Scribe Post: Preparing for the Counting Test

Good day y'all! This is Jann and I'll be your scribe for today.

Reminder: YOU CAN, I repeat, YOU CAN EDIT your DEV projects if you already published your projects.

Morning Class:

1. 7 teams compete in a men's hockey league. If each team plays each other twice, how many games are necessary to complete the league schedule?

Solution:
2. The sum of the seventh row of Pascal's triangle is the same solution to;

Solution:

Since the 7th row in Pascal's Triangle is 64, which is the same as 2^6, we need to find which of these choices have the same answer...
Since (b) is 2^6, the answer is (b).

3. Suppose the last four digits of a telephone number must include at least one repeated digit. How many such numbers are there?

Solution:

4. A multiple choice exam has 20 questions, each with four possible answers, and 10 additional questions, each with five possible answers. How many different answer sheets are possible?

Solution:
5. In a 52 card deck, how many 5 card poker hands are possible that have exactly two pair? one pair?

Solution:That's all we did this morning...

Afternoon Class:

This afternoon, we did our usual group workshops...

Here are the questions:

1.

2.

3.

* the first digit is 9 in (a) because we can't use 0 as a first number. That would make it a 2 digit number instead. (eg. 024)
* the first number is 8 in (b) because we already used a number for the last one to make the digit odd. The next choice is 8 again because this time, we included 0.
* The first solution is for the digit to end in 5, the 2nd solution is for the digit to end in 0. (Both are divisible by 5) For the 2nd solution, the first number is 9 because 0 is in the last digit already, so we don't have to restrict it for the first number.

4.
* In (a), the answer is 7! because we don't care about the order of the books.
* In (b), the answer is 5!3! because if we put the 3 French books in a bag, it is considered as "1". Therefore we have 5 objects to arrange. (5!) After that we remove the 3 books from the bag and multiply 3! to the answer. The 3 books can be arranged in 3 different ways.
* In (c), the question is similar to the "word" problems. Since there are 4 similar objects, they are non-distinguishable.

5.

6.

7.

8.

9.

Reminder: There is no, I repeat, NO test tomorrow because of a multimedia presentation at the gym.

Well... That's all we did for the day. I hope everything was clear. Next scribe is... Grey-M! =D

#### 1 comment:

John D. said...

HA! I was totally right about number 7, I knew it was gonna be the 6th term ahaha! Anyways, good scribe post Jann, you did a good job on explaining the solutions and whatnot.