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School > Data Unit 3 Test > Flashcards

Data Unit 3 Test Flashcards

1
Q

Many counting and probability calculations involve the product of a series of

A

consecutive integers.

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2
Q

You can use … to write such expressions more easily.

A

factorial notation

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3
Q

*n has to be greater than

A

0*

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4
Q

0! =

A

1

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5
Q

(n+3)(n+2)(n+1)… n! = n x (n-1)x(n-2)x(n-3)x…x 3 x 2 x 1 - This expression is read as

A

n factorial

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6
Q

Example: 8! =

A

8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

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7
Q

First, simplify and then

A

find the answer

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8
Q

10!/5! =

A

10 x 9 x 8 x 7 x 6 x 5!/5!
cross out the 5s and solve

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9
Q

(n+3)!/(n+1)! =

A

(n+3)(n+2)(n+1)!/(n+1)!
cross out the (n+1)!s
FOIL
= n^2 + 5n + 6

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10
Q

Cancel two things that are

A

the same on two different sides

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11
Q

n!/(n-2)! = 12

A

n(n-1)(n-2)!/(n-2)! = 12
cross out the (n-2)!s
n(n-1) = 12
FOIL
n^2 - n - 12 = 0
factor
(n-4) (n+3) = 0
n=4 n= -3 but OMIT *can’t be negative)
Therefore, n=4

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12
Q

Tree Diagram:

A

A visual way to organize data so that it is easier to count.
The amount of data must be relatively small or it would be too time-consuming to prepare.
It is most useful to show the connections between objects and to list all of the possible outcomes.

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13
Q

Combinatorics:

A

a branch of mathematics dealing with ideas and methods of counting.

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14
Q

Fundamental Counting Principle or Multiplicative Counting Principle (Product Rule):

A

If one operation can be performed in m ways and for each of these ways a second operation can be performed n ways and for each of these a third operation can be performed p ways,… then all of these can be performed m x n x p x… ways

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15
Q

Additive Counting Principle or Rule of Sum:

A
  • If one mutually exclusive action can occur in m ways and a second can occur in n ways… then there are m + n + p +… ways in which these actions can occur.
  • AKA different cases or examples
  • In each mutually exclusive action, you may be applying the product rule in determining the total number of arrangements for that action
  • AKA You’ll use product rule and then rule of sum for direct questions
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16
Q

Direct Method -

A

when there’s not a lot of cases

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17
Q

“At least” should trigger

A

“cases” so come up with all the different options

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18
Q

Ex. 2 I can make a kabob by using at least 3 different cubes of meat or meat alternative. I have beef, pork chicken, lamb, and tofu. How many different kabobs can I make?

A

Apply product rule for each case

Case 1: 3 Proteins = 5 x 4 x 3 = 60
Case 2: 4 Proteins = 5 x 4 x 3 x 2 = 120
Case 3: 5 Proteins = 5 x 4 x 3 x 2 x 1 = 5! = 120

60 + 120 + 120 = 300

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19
Q

Indirect Method

A
  • Often when there are numerous arrangements to be organized and counted, counting the arrangements that don’t fit the event or action and subtracting it from all the possible arrangements may be necessary.
  • Subtract out the opposite idea
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20
Q

Ex. 3: How many 4-digit numbers (numbers cannot have zero as the first digit) as there that:

a) have no restrictions?

A

9 x 10 x 10 x 10 = 9000
1-9 0-9 0-9 0-9 count zero as an option now

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21
Q

Ex. 3: How many 4-digit numbers (numbers cannot have zero as the first digit) as there that:

b) have no repeated digits?

A

9 x 9 x 8 x 7 = 4536
1-9 add in zero decreases as digit options decrease

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22
Q

Ex. 3: How many 4-digit numbers (numbers cannot have zero as the first digit) as there that:

c) have some repeated digits?

A

Indirect Method = total # of arrangements - no repeated digits
= 9000 - 4536
= 4464

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23
Q

Ex. 4: How many arrangemenisof the letters in the word “hockey” are there if the ‘c’ and the ‘k’ must be kept separate?

A
  • Use indirect method as direct is too complicated (too many cases)
  • There is no way to know how many spaces between the c and the k

Indirect Method = all arrangements of letters in “hockey” - when c and k are together
= 6! - 5! (treat CK as one item) x 2! (CK or KC)
= 720 - (120 x 2)
= 720 - 240
= 480

24
Q

Permutations

A

are possible groupings or arrangements with respect to order.

25
Q

P(n,n) =

A

n!

26
Q

P(n,r) =

A

n!/(n-r)!

27
Q

P(n,r) can also be written as

A

nPr “n pick r”

28
Q

HOW MANY WAYS =

A

USE N PICK R

29
Q

Ex. 1
A volunteer group has 24 members. They wish to elect a chairperson, a vice-chairperson, and a
treasurer. In how many ways can this be done?

A

P(24,3) =
24!/(24-3)! =
24!/21! =
24x23x22x21!/21! =
cross out the 21!s =
24x23x22 =
12144 ways

30
Q

Ex. 3
In how many ways can 8 songs be burned on a CD if there are 12 songs in my library?
songs are distinct anf order matters

A

P(12,8) =
12!/4! =
19,958,400

31
Q

Ex. 4
At a girl’s preschool, there are 9 girls, 3 of which are sisters. In how many ways can:

a) All 9 girls arranged in a line

A

“9 pick 9”
P(9,9)
= 9!
= 362880

32
Q

Ex. 4
At a girl’s preschool, there are 9 girls, 3 of which are sisters. In how many ways can:

b) Six girls be selected and arranged in a line:

A

P(9,6) =
9!/3! =
60480

33
Q

Ex. 4
At a girl’s preschool, there are 9 girls, 3 of which are sisters. In how many ways can:

c) All girls arranged with the tallest in the middie:

A

8 7 6 5 1 4 3 2 1
P(8,8) =
8! =
40320

Always start with the restrictions
Place the restricitins then arrange the rest

34
Q

Ex. 4
At a girl’s preschool, there are 9 girls, 3 of which are sisters. In how many ways can:

d) All girls arranged if the sisters must remain together

A

Treat the sisters as 1 item
= 7 items to arrange = 7!
But there are also P(3,3) ways to arrange the sisters
= P(7,7) x P(3,3)
= 7! X 3!
= 30240

35
Q

Ex. 4
At a girl’s preschool, there are 9 girls, 3 of which are sisters. In how many ways can:

e) All girls arranged if the sisters must be apart

A

Must use this diagram
^^^^^

_ = non-sisters (6)
^ = where sisters can be placed, ensuring all sisters are separated

non-sisters (_) x sisters (^)
6! X P(7,3)
= 6! X 7!/4!
= 6! X 210
= 151,200

36
Q

Permutations with Repetition

A

In general, the number of different arrangements of n objects with a alike of one kind, b alike of another kind, … is: n!/a!b!

37
Q

“UNIQUE” MEANS USE

A

REPETITION FORMULA

38
Q

How many arrangements are there of the letters in the word MISSISSIPPI if,

a) if there are no restrictions?

A

11 letters
= P(11,11) = 11! = 344 16200

39
Q

How many arrangements are there of the letters in the word MISSISSIPPI if,

b) if the permutations are unique?

A

Must get rid of duplicates
4S
4I
2P
= 11!/2!4!4! = 34650

40
Q

How many arrangements are there of the letters in the word MISSISSIPPI if,

c) if the first letter must be an m?

A

Put m in the first position and arrange the remaining letters

1 <____10!______>
M

= 10!/2!4!4!
= 3150

41
Q

How many arrangements are there of the letters in the word MISSISSIPPI if,

d) the last letter must be a vowel?

A

Take one i (only vowel) and put in in the last position

<____10!______>1
I

= 10!/2!4!3! (only 3 Is left)
= 12600

42
Q

How many arrangements are there of the letters in the word MISSISSIPPI if,

e) the last letter must be a consonant?

A

Direct method add up cases

Case 1: ends in ‘m’ = 10!/2!4!4! =3150
Case 2: ends in ‘s’ = 10!/2!3!4! = 12600
Case 3: ends in ‘I’ = 10!/1!4!4! = 6300

Total = 3150 + 12600 + 6300
= 22050

“Rule of sum”

43
Q

In cases where things need to be kept apart but they also have to be unique because some of the items are the same…

A
  • Combine the ^^^ method with the
  • Unique method
44
Q

c) are unique and have all the vowels apart

A

(_) x (^)
5! P(6,4) treat as an item and how many vowels total
= 5!/2!2! (because those are the repeated consonants) x 6!/4!2! (because those are repeated vowels)
= 120/4 x 6x5x4!/4!2!
= cross out the 4!s
= 120/4 x 6x5/2!
= 120/4 x 30/2
= 30 x 15
= 450

45
Q

Pascals Triangle

A

Each number is the sum of the two numbers directly above it to the right and left

46
Q

Row sum property:

A

The sum of the terms in any row n is 2^n

47
Q

e.g. The sum of the terms in row 5 is

A

2^5 or 32.
Row 5 = 1 + 5 + 10 + 10 + 5 + 1 = 32

48
Q

Ex. 2^n = 32768 = n =

A

15th row

49
Q

Ex. 1: In the arrangement of the word “SERIES” given below, starting from the top, we
proceed to the row below by moving diagonally to the right or the left. How many paths
can spell series?

a) Triangle

A
  • Put 1s on the top and sides of the s and down the sides
  • STOP ADDING BEFORE THE LAST ROW OF LETTERS
  • Add in the triangle
50
Q

Ex. 1: In the arrangement of the word “SERIES” given below, starting from the top, we
proceed to the row below by moving diagonally to the right or the left. How many paths
can spell series?

b) Diamond

A
  • Put 1s on the top and sides of the s until the letters BEFORE the peak row of the letters
  • STOP ADDING BEFORE THE LAST ROW OF LETTERS
  • Add in the diamond
51
Q

Ex. 2: How many routes are there from home to school? (You can only move south or
east at each intersection).

A
  • Down or right
  • Do the 1s down the square and to the right side
  • Add in the bottom right
  • The last square if the answer
52
Q

Ex. 3: If you can only travel north or east, determine the number of pathways from A to
B in the following street arrangement?

A
  • Up and right
  • Put ones down the sides
  • Add in the upper right
  • The last square is the answer, if it gets cut off into a new square, use the last number as the new “1” and write it down the sides
  • The last square is the answer
53
Q

Ex. 4: How many paths to the top of the checkerboard?
(You cannot move the spot marked ‘X’)

A
  • Can only move diagonally
  • Put the 1s into all the spots diagonal from the pieces
  • Add as you go
  • When the ones end use the last diagonal number to expand
  • When there’s an x “x + 5” is still 5 etc
  • Add the last squares for the answer
54
Q

Q: How would the total number of paths change if you could jump over the X?

A

Bring the last number up the x and add it with the number beside the x

55
Q

How to factor

A

Add to B and multiply to C

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