## November 30, 2006

### scribe post for today

Another winter day had past and we had an uneasy day due to the test we had this afternoon. In our first period pre-calculus, we had a pretest in relation to our actual examination where we did individual worksheets then group work and then a two and a half minute spy time. After that Mr. K revealed the answers for the following questions:

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's schedule?

(a) 42
(b) 21
(c) 84
(d) 52

The answer is (a) 42 because the solution is "2*(7C2)" where 7 is the number of teams and there must be at least first two teams to compete then multiplied by 2 because they must compete with each other twice.

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

(a) The number of solutions a student could get if they guessed on six questions of a multiple choice exam, each question having four answers.
(b) The number of solutions a student could get if they guessed on six questions of a true false exam.
(c) The number of solutions a student could get if they guessed on seven questions of a multiple choice exam, each question having four answers.
(d)
The number of solutions a student could get if they guessed on seven questions of a true false exam.

The answer is (b) because in six questions there are two choices for each, which is either true or false. Therefore, 2*2*2*2*2*2 or 2^6 equals 264.

Solution:
_ _ _ _ _ _
2 2 2 2 2 2 = 264

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

The answer is 4,960 because there are 10^4 or 10,000 ways of having last four digit numbers if there is no restriction and there are 10*9*8*7 or 5040 ways of having last four digit numbers with restriction. In order to solve for the number of ways to have last four numbers with at least one repeated number, we have to subtract 5,040 from 10,000. As a result, we got 4,960.

Solution: 10,000-5,040= 4,960

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?

There are (4^20)*(5^10) or 1.073741824 x 10^19 possible answer sheets because in 20 questions there are four possible answers, 4^20, for each and in 10 questions there are five possible answers, 5^10, for each. In order to get the answer we have to multiply both (4^20) and (5^10) with each other because we are looking for the differenr number of answer sheet possible.

Solution:
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4= 4^20

_ _ _ _ _ _ _ _ _ _
5 5 5 5 5 5 5 5 5 5= 5^10

(4^20)*(5^10)= 1.073741824 x 10^19

5) In a 52 card deck, how many 5 card poker hands that have exactly two pair? one pair?

in two pairs: There are 123,552 possible cards with exactly two pairs.

Solution: 13C2* 4C2* 4C2* 11C1* 4C1= 123,552

where:
13 is the number of face values from ace to king
4 is the number of suits per face value (hearts, clubs, diamonds, and spades)

13C2 is the equation for a person to choose 2 face values out of 13. Two 4C2 are given because when a person have already chosen two face values, he will then choose one pair of number from four suits, which are the hearts, diamonds, clubs, and spades and another pair from the other suit. 11C1 is the equation for the remaining face values where the person has to choose only one from it. 4C1 is given because after the selection from the remaining 11 face values, the person has to choose one number from any suits of the same number.

in one pair: There are 1,098,240 possible cards with exactly one pair.

Solution: 13C1* 4C2* 12C3* 4C1* 4C1* 4C1 or 13C1* 4C2* 12C3* (4C1)^3= 1,098,240

13C1 is the equation for the 13 face values where a person has to choose only one from it. 4C2 is given because after he has chosen the face value, he has to choose one pair from four different suits of the same number. 12C3 is the equation for the remaining face values where a person has to choose three from it. Three 4C1 are given because in different face values there are four suits each and he has to choose only one number from the four suits of the same number resulting into having three different numbers.

In our second period pre-calculus, we started answering our Combinatorics test where all of us are hoping for the best test results.

That's all for now guys.
The next scribe is Allen.
Goodluck to all of us.....
God bless you all..........

------(",) daphne.......