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School > Data Unit 4 Test > Flashcards

Data Unit 4 Test Flashcards

1
Q

S

A

Universal Set (this represents everybody) Written in the top left corner

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2
Q

Write leftovers in the bottom right corner of the

A

Venn diagram

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3
Q

Always draw a rectangle around the

A

Venn diagram

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4
Q

When calculating the total of a venn diagram

A

add all the numbers together

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5
Q

When calculating the number of students in a specific class,

A

add all numbers in a circle

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6
Q

n =

A

number

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7
Q

Where U is “the

A

union of” (or the word “or” is used > implies addition)

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8
Q

Where ∩ “the

A

intersection of” (or the word “and” is used > implies multiplication)

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9
Q

Principle of Inclusion & Exclusion for 2 Sets Formula:

A

n(A U B) = n(A) + n(B) - n(A ∩ B)

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10
Q

At least one in Venn Digrams -

A

One option, the other option, or both options

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11
Q

When using Venn DIagrams, ALWAYS START WITH THE

A

OVERLAP

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12
Q

At least

A

Union U

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13
Q

More than one =

A

Add up all the overlaps - with a Venn diagram it can’t be taken straight from the data or students will be counted twice

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14
Q

Principle of Inclusion & Exclusion for 3 Sets Formula:

A

n(C U F U D) = n(C) + n(F) + n(D) - n(C∩F) - (C∩D) - n(F∩D) + n(C∩F∩D)

  • Overlaps are subtracted and added back in
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15
Q

Order is

A

not important

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16
Q

It is just the combination of things,

A

not in any specific order

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17
Q

Combination formula

A

C( n , r ) = n! /( n - r )! r!

-It is the permutation formula divided by duplicates (removing order, dividing out the permutations)

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18
Q

Ex. Evaluate a) (10,2) b) C(10,8)

A

45 and 45 are the answers - You can pick 8 for a committee and leave 2 behind in the same # of ways

Same options for selecting r and not selecting r.

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19
Q

Ex. 1
In how many ways can 4 cards be dealt from a regular deck of 52 playing cards if:

a) No restrictions

A

“52 choose 4” C(52,4) = 270,725

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20
Q

Ex. 1
In how many ways can 4 cards be dealt from a regular deck of 52 playing cards if:

b) One card must be the ace of spades

A

Ace: C(1,1) Spades: C(51,3) = 20825
1 + 51 = 52, 1 + 3 = 4

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21
Q

Ex. 1
In how many ways can 4 cards be dealt from a regular deck of 52 playing cards if:

c) Exactly one card must be a 2

A

2’s: C(4,1) Not 2: C(48,3) = 69184
52 = 4 + 48, 4 = 1 + 3

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22
Q

Ex. 1
In how many ways can 4 cards be dealt from a regular deck of 52 playing cards if:

d) At least one card must be a 2

A

Direct
C(4,1) C(48,3) + C(4,2) C(48,2) + C(4,3) C(48,1) + C(4,4) C(48,0)
one 2 two 2’s three 2’s four 2’s
= 4(17296) + 6(128) + 9(48) + 1(1)
= 69184 + 6768 + 192 + 1
= 76145

Indirect
All possible outcomes - no 2’s
C(52,4) - C(48,4) = 270725 - 194580
= 76145

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23
Q

Ex. 2
How many bridge hands (13 cards) contain:

a) Exactly 8 spades?

A

C(13,8) C(39,5)
Spades Rest
= !287(575757)
= 740,999,259

8 + 5 =13 card hand
13 + 39 = 52 cards, 13 spades, 39 not spades

24
Q

Ex. 2
How many bridge hands (13 cards) contain:

b) Exactly 8 cards of any suit?

A

Almost the same as a) but we must account for all suits not just spades, therefore C(4,1) ways to choose a suit
= C(4,1) C(13,8) C(39,5)

Choose a suit, 8 from that suit, 5 leftovers

think of hierarchy, what has to happen first? > pick a suit first, then cards

25
Q

Opening Exercise:
In how many ways can a basketball team of 9 players be selected from 15 prospects (3 of which are returnees from last year) if:

a) no restrictions

A

C(15,9) = 5005

26
Q

Opening Exercise:
In how many ways can a basketball team of 9 players be selected from 15 prospects (3 of which are returnees from last year) if:

b) there must be exactly one returnee

A

hierarchy returnees > rest
C(3,1) C(12,8) = 1485

27
Q

Opening Exercise:
In how many ways can a basketball team of 9 players be selected from 15 prospects (3 of which are returnees from last year) if:

c) There must be at least one returnee

A

Solution 1: Indirectly
All options - no returnees
C(15,9) - C(12,9) = 4785

Solution 2: Directly
1 returnee + 2 returnees + 3 returnees
C(3,1) C(12,8) + C(3,2) C(12,7) + C(3,3) C(12,6) = 4785

28
Q

some =

A

at least one

29
Q

Product Rule with 2^ Approach ->

A

for anything that has 2 options (win or lose, heads or tails, true or false)

30
Q

The formula for the product rule when there are 2 options is

A

2^n - 1.

31
Q

In situations where you are asked to count the number of ways that items can be selected in different quantities or subsets, it can be done using either the product rule or the rule of sum in which the direct or indirect method may be used. When words like “some” and “at least” are used, it means having to select

A

one or more items and each quantity represents a mutually exclusive subset or combination.

32
Q

Example 1. A tray of candies contains 1 red, 1 blue, 1 yellow, 1 green,1 purple, and 1 black.
In how many ways can some candies be pulled from the tray?

Solution 1: Product Rule Approach -> for anything that has 2 options (win or lose, heads or tails, true or false)

A

Thus, the total number of combinations containing at least one item chosen from n distinct items is:
2^6 - 1
= 64 - 1
= 63

33
Q

Order doesn’t matter since we are not lining them up. For each candy, you have 2 choices: to take it from the tray or not. Therefore, with six pieces of candy, there are 2x2x2x2x2x2 = 2^6 = 64 total combinations. But one of the combinations is to not pull any candy at all, called the null set, which goes against the question of having some candies pulled.

A

Therefore there are
2^6 - 1 = 63 ways.

34
Q

You have to take away the null case C(6,0)

A

here we choose zero candies.

35
Q

Example 1. A tray of candies contains 1 red, 1 blue, 1 yellow, 1 green,1 purple, and 1 black.
In how many ways can some candies be pulled from the tray?

Solution 2: Rule of Sum Approach

A

6 cases: C(6,1) + … C(6,6) = 63

36
Q

Example 2. How many committees of 8 or less can be formed from any group of 10 people?

Solution 1: Indirect

A

= all possible subsets - (10 people + 9 people + 0 people)
= 2^10 - (C(10,10) + C(10,9) + C(10,0))
= 1012

37
Q

Example 2. How many committees of 8 or less can be formed from any group of 10 people?

Solution 2: Direct Method: all cases

A

= C(10,1) + C(10,2) …. C(10,8)
= 1012

38
Q

Example 3. A basket of fruit contains 3 apples, 2 mangos, 5 plums, and 9 bananas. In how many ways can some fruit be selected from the basket?

Some = ?
We don’t need ? of each fruit

A

Some = at least one
We don’t need 1 of each fruit

39
Q

Here, there is repetition that you have to deal with as not all the items are distinct. Furthermore,
Fruit picking “some fruit” again implies counting all subsets of the fruits with at least one fruit being selected. Unlike permutations, we cannot simply divide each subset by the factorial of the like items since order does not matter. In this situation, when it comes to selecting all, some, or none of p alike items, you have p +1 choices. Then, if you have successive choices of different kinds of items, you would apply the product rule. Remember, the null set of selecting no items will be counted and must be

A

considered in evaluating our solution.

40
Q

If some items are alike and if at least one item is to be chosen, then the total number of selections

A

from p alike items, q alike items, r alike items, and so on is:

41
Q

The formula for DIFFERENT items WITH MANY being selected or not:

A

(p+1)(q+1)(r+1)…. -1,

42
Q

(p+1)(q+1)(r+1)…. -1,

A

1 represents the null set.

43
Q

Example 3. A basket of fruit contains 3 apples, 2 mangos, 5 plums, and 9 bananas. In how many ways can some fruit be selected from the basket?

Solution: Here, the order is still not important since we are not lining them up; however, some of the objects are alike. With our apples, we can select 1, 2, 3, or 0 apples which are 4 options. For our mangos, we can select 0, 1, or 2, and so on for the other fruits.

A

Therefore, the total ways to select some fruit is:
(3+1)(2+1)(5+1)(9+1) - 1 = 719

Apples, mangoes, plums, bananas, null set - can’t account for none chosen the case where we don’t pick ANYTHING has to be subtracted

44
Q

CARDS
52 total

A

26 Black 26 Red

45
Q

CARDS
4 Suits

A

13 Black Spades
13 Black Clubs
13 Red Hearts
13 Red Diamonds

46
Q

CARDS
12 Face Cards

A

3 J Q K for each suit

3x4 = 12

47
Q

Permutation Formula

A

P(n,r) = n!/(n-r)!

48
Q

Combination formula

A

C(n,r) = n!/r!(n-r)!

49
Q

How to solve for N

A

Expand the n! to the (n-r) in the denominator
Cross out anything on both sides (n)
Divide one side by any multiple on the other
FOIL
Make one side equal to zero
Factor
OMIT

50
Q

IN HOW MANY WAYS CAN A COMMITTEE OF 4 PEOPLE BE CHOSEN FROM 6 MARRIED COUPLES IF NO 2 PEOPLE FROM THE SAME COUPLE CAN BE ON THE COMMITTEE?

A

C(6,4)C(2,1)C(2,1)C(2,1)C(2,1)
C(6,4) 2^4
15 x 16
240

51
Q

AT LEAST TRIGGERS

A

CASESSSSSS

52
Q

when “at least” cases, when you have cases you

A

ADDDDDD

53
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

54
Q

how to factor

A

add to b multiply to c

55
Q

cancel two things that are

A

the same on both sides

56
Q

2n!/(n-3)! = 84n

A
  1. expand the numerator to cross out the denominator
    2(n)(n-1)(n-2)(n-3)!/(n-3)! = 84n
    2(n)(n-1)(n-2) = 84n
  2. cross out the n’s as they are the same on both sides
    2(n-1)(n-2) = 84
  3. divide 84 by 2
    (n-1)(n-2) = 42
  4. FOIL
    n^2 - 3n + 2 = 42
  5. make it equal to 0
    n^2 - 3n - 40 = 0
  6. factor (adds to b multiplies to c)
    (n-8)(n+5) = 0
    n=8 n= -5
  7. OMIT -5 (can’t be negative)
    therefore, n=8

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