### Scribe Post: the pre-test

Ladies and gentleman, tonight we bring you the ever so awsome scribe of the night, teddie!!! yay!!!

Well today was a one period day and we did the pre-test for the main event the merciless, breath-taking . . . . COUNTING TEST. Well I will make this scribe more of a review or tool to help you guys study for this up coming test. So, first and foremost I will review the pre-test we did today.

PRE-TEST ON COMBINATORICS

(1) How many different committees of 2 people can be selected from 5 people?

(A) 5! / 2! (B) 5! / 3! (C) 5! / 3! 2! (D) 5! / 2(3)

Well let me show you what we look at or for in the question and from there we will find the answer. The total amount of people you are working with is 5. And the question is aking for 2 of any of the 5. Knowing that it's just a committee, we know that the order we choose them does not matter. So just from knowing that, we know that we don't use the pick formula. Therefore, the formula we use for this case is the choose formula because, order doesn't matter. the formula is:

_{n}C

_{r}= n! / (n - r)! r!

in this question n = 5

and r = 2

5 choose 2 simple right?

solving:

5! / (5 - 2)! 2!

5! / 3! 2! [this is choice C]

(2) Determining the 5th term in the expansion of (x - 1/2y)

^{7}.

(A) 35/8 x

^{4}y

^{3}(B) 35/16 x

^{3}y

^{4}(C) - 35/8 x

^{4}y

^{3}(D) - 35/16 x

^{3}y

^{4}

This question is an example that we use the Binomial Theorum.

_{7}C

_{4}a

^{3}b

^{4}

let a = x , b = -1/2y

_{7}C

_{4}(x)

^{3}(-1/2y)

^{4}

7! / (7 - 4)! 4! x

^{3}1 /16y

^{4}

7! / 3! 4! x

^{3}1 /16y

^{4}

7*6*5*

^{3}1 /16y

^{4}

35 x

^{3}1 /16y

^{4}

35 / 16 x

^{3}y

^{4}[this is choice B]

REMEMBER: when it asks for a certain term for example the 5th term, the number we replace for r in the equation is one less than the term, in this case 5 - 1 = 4.

(3) Moving only to the right or down, how many different paths exist to get from point A to point B?

(A) 22 (B) 60 (C) 120 (D) 144

At first this question looks crazy confusing but if we use what we have learned some useful tools in this unit that can help to tackle this problem. First we break up the question. we break apart the 3 boxes and solve each and then we will use the Fundamental Principle of Counting to bring it together. Looking at each box, we can treat each time we travel across one little box length and a letter. For each time we move right one box, we will call it R. And for each time we move down on box we will call it D.

Looking at the first box (3 x 2 squares) we know that in order to get from point A to point X we must move right 3 R units and down 2 D units. Therefore we create the word RRRDD. Now we turned this question into a Permutation of Non-Distinguishable objects. The 3 R's and the 2 D's are both non-distinguishable from themselves. And all the possible ways to rearrange this word is 5!. Therefore we come up with this:

5! / 3! 2!

5*4*

20 / 2

10

We do the same for the other boxes too. For the 2 by 2 box, we create the word RRDD, going from X to Y. And all the possible ways to rearange this word is 4!. So can solve this box.

4! / 2! 2!

4*3*

12 / 2

6

For the final 1 by 1 box. we can create the word RD, goin from Y to B All the possible ways to rearange this word is 2! So we can solve this box.

2! / 1! 1!

2*

2

Now we apply the Fundamental Principle of Counting.

10 * 6 * 2

120 [this is choice C]

(4) How many differnet 10-digit telephone numbers can be created if the area code (first three digits) is 250 or 604 and the exchange (second three digits) is 531, 535 or 540?

(A) 50 000 (B) 60 000 (C) 80 000 (D) 90 000

The way you solve this takes a little more explaination. The way I look at this question, I see the numbers in a tree diagram.

As you can see in the diagram above, each area code is the main numbers to branch off. Then connected to the area codes are 3 of the different exchanges each. And for each exchange we know that the last 4 digits can contain all possible single digit numbers which is 10 each. Basically the restrictions are, there can only be 2 possible area codes and 3 possible exchnanges. Therefore we can solve it using the choose formula.

_{2}C

_{1}*

_{3}C

_{1}* (

_{10}C

_{1})

^{4}

2 * 3 * 10 000

60 000 [this is choice B]

(5) A "palindrome" is a number that reads the same forwards and backwards, such as 121, 12321, or 4554. Determine the number of 7 - digit palindromic phone numbers that can be created if the only restriction is that 0 cannot appear as teh forst or third digit.

(A) 810 (B) 8 100 (C) 9 000 (D) 8 100 000

Well first of all we know we must find seven numbers to make up a phone number.

_ _ _ _ _ _ _

And in phone numbers, we can use all possible single digit numbers for each of the seven numbers.

10 * 10 * 10 * 10 * 10 * 10 * 10

We also know that the whole number is palindromic which means, whichever numbers thats is chosen on one side has to be the same on the other.

!0 * 10 * 10 * 10 * 1 * 1 * 1

But there are restrictions to this set of numbers. The first and the third number of the phone number cannot be zero. Taking out one choice of numbers from the first and third slots.

9 * 10 * 9 * 10 * 1 * 1 * 1

8 100 [this is choice B]

Long answer Questions

I have 5 different math books and 6 different physics books.

(a) In how many ways can 2 math books and 3 physics books be arranged on a book shelf if the math books have to be together and the physics books have to be together?

Well from the 5 math books we only want 2 and from the 6 physics books we only want 3. And you want to rearrange the books on a shelf either with all the math books first then all the physics books or the other way around, therefore have only 2 ways to do that. The way you can rearrange the math books together is 2! because they are different and distinguishable. The way you can rearrange the physics books together is 3!. Using all this information we can come up with the answer by using the Choose Formula combined with the Fundamental Principle of Counting.

_{5}C

_{2}*

_{6}C

_{3}* 2 * 2! * 3!

5! / (5 - 2)! 2! * 6! / (6 - 3)! 3! * 2 * 2 * 6

5! / 3! 2! * 6! 3! 3! * 24

10 * 20 * 24

4800

(b) In how many ways can 5 differnt books be arranged on the shelf if at least 2 of them have to be math books?

Our first reaction probably is to find all the possibilities that the math books can be rearranged when we have 2, 3, 4 and 5. Sure that will give us the answer but there is an easier way. We can find the total possiblilities with no restrictions, subtract the possibilities for having no math books and subtracting the possiblilites of have just one math book. In other words we find the compliment. Remember that we have 5 math books and 6 physics books.

TOTAL (no restrictions)

_{11}C

_{5}* 5!

11! / 6! 5! * 120

462 * 120

55440

NO MATH BOOKS (all physics)

_{6}C

_{5}* 5!

6! / 5! * 5!

6 * 120

720

1 MATH BOOK

_{5}C

_{1}*

_{6}C

_{4}* 5!

5 * 6! / 2! 4! * 5!

5 * 15 * 120

9000

TOTAL (no restrictions) - TOTAL (no math books) - TOTAL (just 1 math book)

55440 - 720 - 9000

45720

Well that's all ladies and gentlemen that is the whole pre-test reviewage for ya. And tomorrows scribe is . . . . the one and the only Manny Fresh lol. Laters!