Combinations and Permutations

What's the Difference? If the order doesn't matter, it is a Combination If the order does matter it is a Permutation Combinations: Drawing or combination of quantity without order of objects or things. Example: "My fruit cocktail is a combination of mangoes, papayas and bananas". I don't really care what order the fruits are in, they could also be "bananas, papayas and mangoes" or "papayas, mangoes and bananas" etc., its the same fruit cocktail. Permutations: Drawing or combination of quantity with order of objects or things. Example: "The combination to the emergency number is 911". Now I do care about the order. "191" won't work, nor will "119". It has to be exactly 9 -1 -1. Notice: in Mathematics: If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation. Remember Permutation = Position! There are also two types of combinations (remember the order does not matter now): Repetition is Allowed: such as coins in your pocket (4, 4, 4.10, 10 etc.) No Repetition: such as lottery numbers (2.14,15.27,30.33) Combinations with repetition coming soon ... Combinations without Repetition That's how lotteries work. The numbers are drawn one at a time, and if you have the lucky numbers (no matter what order) you win! Formula and different Notations c(n, r)  =  ncr  =  ncr  =  ( n r )  =  n! r!(n − r)! with: c the combination n the total number of items and r the number taken/chosen at one time (once) Here some examples: The oder doesn't matter!!! We abandon the order of objects. We draw three items (numbered 1, 2, 3) in the bag of 10 different objects. How many opportunities, regardless the arrival sequence of objects do we have? The answer: 10c3  =  10! 3!(10 − 3)!  =  10 · 9 · 8 · 7! 3! · 7!  =  120 Choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations. The answer: 16c3  =  16! 3!(16 − 3)!  =  16 · 15 · 14 · 13! 3! · 13!  =  560 16c13  =  16! 13!(16 − 13)!  =  16 · 15 · 14 · 13! 13! · 3!  =  560 Top | Permutations | Go to Exercises

Brüche Multiplizieren bzw. Dividieren

Aufgabe 7: Multipliziere die gemischten Zahlen. a)  1 3 5 · 2 1 4 ;      b)  2 3 · 3 1 2 ;      c)  2 3 5 · 3 ;      d)  3 1 3 · 4 1 2 ;      e)  1 1 2 · 3 4 5 ;  Aufgabe 2xtra: Wähle den unechten Bruch. 1.  1 3 4                2.  1 4 5                 3.  2 3 10                   4.  5 3 4                   5.  5 5 8 7/4 9/4      9/5 10/5     15/10 23/10        23/4 12/4        18/8 45/8

Richtige Antworte

Übungen zu Brüche

Aufgabe 1: Trage den unechten Bruch ein. a)   2 1 3  =       b)   1 3 5  =       c)   3 4 5  =       d)   5 1 10  =          a)   7 3               b)   8 5             c)   19 5                d)   51 10 Aufgabe 2: Trage die gemischte Zahl ein. a)  43 10   =   b)  39 10   =   c)  11 2   =             a)   4 3 10           b)   3 9 10           c)   5 1 2 Aufgabe 3: Addiere und kürze soweit wie möglich. a)  2 5  +  3 5 ;  b)  11 14  +  5 14 ;  c)  1 3  +  1 4 ;  d)  1 8  +  3 8 ;  e)  3 4  +  1 6 ;  f)  1 2  +  5 8 zu den Lösungen3 Aufgabe 4: Subtrahiere und kürze soweit wie möglich. a)  6 7 − 3 7 ;    b)  7 8 − 5 6 ;    c)  3 4 − 1 4 ;    d)  5 7 − 10 14 ;    e)  5 7 − 1 2 ;    f)  3 4 − 4 12 ;    g)  7 10 − 3 8 h)  2 3 − 5 9 zu den Lösungen4 Aufgabe 5: Addiere die gemischten Zahlen. a)  4 14 17  +  1 17 ;   b)  2 7 11  +  4 1 17 ;   c)  5 8 17  +  3 3 17 ;   d)  1 1 2  +  1 3 e)  8 1 3  +  4 1 2 ;   f)  7 4 5  +  8 1 2 zu den Lösungen5 Top ⤴ Aufgabe 6: Subtrahiere die gemischten Zahlen. a)  8 3 4 − 5 2 3 ;   b)  5 5 8 − 3 4 5 ;   c)  3 1 2 − 3 8 ;   d)  7 3 4 − 1 2 ;   e)  4 5 9 − 2 4 7    f)  12 2 3 − 7 5 6 ;   g)  3 3 5 − 1 6 11 ;   h)  2 1 2 − 1 4 5 zu den Lösungen6 Weitere Brüche ⇨ Lösungen Lösungen3 a)  1 ;  b)  1 1 7 ;  c)  7 12 ;  d)  1 2 ;  e)  11 12 ;  f)  1 1 8 Lösungen4 a)  3 7 ;  b)  1 24 ;  c)  1 2 ;  d)  0 ;  e)  3 14 ;  f)  5 12 ;  g)  13 40 ;  h)  1 9   Lösungen5 a)  4 15 17 ;  b)  6 130 187 ;  c)  8 11 17 ;  d)  1 5 6 ;  e)  12 5 6 ;  f)  16 3 10   Lösungen6 a)   3 1 12 ;   b)  − 7 40 ;   c)   3 1 8 ;   d)   7 1 4 ;   e)   1 62 63 ;   f)   4 5 6    g)   2 3 55 ;   h)   7 10 ;   zum Kernunterricht

Brüche addieren bzw. Subtrahieren

Gleichnamige Brüche Addieren bzw. Subtrahieren

Gleichnamige Brüche:

• haben den gleichen Nenner.
• werden miteinander addiert bzw. subtrahiert, indem man die Zähler miteinander addiert bzw. subtrahiert und den Nenner beibehält.

Beispiele

3 2  −  1 2  =  3 − 1 2  =  2 2  =  1,

8 12  +  4 12  =  12 12  =  1,

1 8

+

3 8

=

1 + 3 8

=

4 8

=

1 2

Ungleichnamige Brüche Addieren Subtrahireren

Ungleichnamige Brüche werden in zwei Schritten addiert bzw. Subtrahiert:

• auf den Hauptnenner (Gleicher Nenner) bringen
• und die Zähler miteinander addieren bzw. subtrahiren genauso wie bei Gleichnamige Brüche.

Beispiele

1 3  +  1 4  =  4 12  +  3 12  =  4 + 3 12  =  7 12

,

3 4  −  1 2  =  6 8  −  4 8  =  6 − 4 8  =  2 8  =  1 4

Umwandlung:
Unechte Brüche in gemischte Zahlen - Gemischte Zahlen in Unechte Brüche

Unechter Bruch: Ein unechter Bruch ist ein Bruch, dessen Zähler größer ist als sein Nenner. z.B.: 19 6

Gemischte Zahl: Eine gemischte Zahl besteht aus einer ganzen Zahl und einem Bruch. z.B.: 3 2 5

Unechte Brüche in gemischte Zahlen umwandeln

Gegeben:

17 5

Den Zähler wird durch den Nenner normal dividiert:
17 : 5 = 3, Rest = 2

• Das ganzzahlige Ergebnis gibt die Anzahl der Ganzen an;

z.B.:

3

2 5

(3 ist das ganzzahlige Ergebnis).

• Der Rest gibt den Zähler des Restbruches an;

z.B.:

3

2 5

(2 ist der Rest).

• Der Nenner des Restbruches ist der Nenner der ursprünglichen Bruch.

z.B.:

3

2 5

(5 ist der Nenner).

Gemischte Zahlen in Unechte Brüche

Gegeben:

8 1 3

⇒	(3 · 8) + 1 3 = 24 + 1 3 = 25 3

zu den Übungen

Exercises Coordinate Geometry 2

Solution to 7

Exercises Coordinate Geometry 1

...

Solution to 5

(a) To show that all four sides are equal in length, we need to calculate their distances (length), using the Distance Formula.

d = √(changing in x)² + (changing in y)² d = √(x2 − x1)² + (y2 − y1)²

1. Distance between A(− 3, 2) and B(4, 3)
d_AB = √(4 − (− 3))² + (3 − 2)² = √7² + 1² = √50 = √2 · 5² = 5√2


2. Distance between B(4, 3) and C(9, − 2)
d_BC = √(9 − 4)² + (− 2 − 3)² = √5² + (−5)² = √50 = √2 · 5² = 5√2


3. Distance between C(9, − 2) and D(2, − 3)
d_CD = √(2 − 9)² + (− 3 − (− 2))² = √(−7)² + (−1)² = √50 = √2 · 5² = 5√2


4. Distance between D(2, − 3) and A(− 3, 2)
d_DA = √(2 − (−3))² + (− 3 − 2)² = √5² + (−5)² = √50 = √2 · 5² = 5√2

(b) As you can notice:
      - All for sides are equal in length
      - Opposite sides are parallel
but
      - The four angles are not right
      - The diagonals are not congruent

Therefore, the figure is still a quadrilateral, not a square, but a parallelogram.

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Solution to 6

(a) First solve the line eqaution 3x + 4y = 16 for y:
      that way you'll come to l₁: y = ⁻³/₄·x + 4, with the slope m₁ = ⁻³/₄.

      If l₂ passes through P and perpendicular to l₁, it means that the slope of m₂ ist negative       reciprocal to the slope of m₁
      ⇒ m₁ = ⁻¹/_m2 ⇔ m₂ = ⁻¹/_m1 = ⁻¹/_⁻³/4 = ⁴/₃, m₂ = ⁴/₃
      l₂ passes through P
      ⇒ 5 = ⁴/₃ − 7 + b, with b to be found!
      5 = ²⁸/₃ + b and b = ⁻¹³/₃

      Therefore, the equation of l₂: y = ⁴/₃·x − ¹³/₃

(b) Point of intersection of l₁ and l₂:
      this means l₁ = l₂
      ⇒ ⁻³/₄·x + 4 = ⁴/₃·x − ¹³/₃, solve for x:
      ⇒ x = 4
      then plot x into either l₁ or l₂:
      ⇒ y = ⁻³/₄·4 + 4 =
      In conclusion, the points l₁ and l₂ intersect at the point (4, 1)

(c) The perpendicular distance of P from the line l₁ represents the Hypotenuse of the slope       with the values x = 3 and y = 4
      therefore, we will use the Pythagoras formula for right triangle to find out the distance      concerned:

      so, d² = 3² + 4²    |   √ (square root it)
      ⇒ d = √3² + 4² = √5²
      ⇒ d = 5

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continue...

Exercises Coordinate Geometry

Solutions

bottom ↓

2. y = 4x − 8 meets the x-axis means that y = 0
by solving 4x − 8 = 0, it appears that x = 2
⇒ the line meets the x-axis at the point A(2, 0)

We'll now find the equation of the line with gradient 3, passing through the point A
The gradient m = ^{change in y}/_{change in x} = ³/₁, which means the line passes through another point B(1, 3) with A(2, 0) as origin. (see Plot 1)



Plot 1




3. y = − 2x + 8 meets the y-axis means that x = 0
⇒ y = − 2(0) + 8 and y = 8
The point met is B(0, 8)
The other line passing, with gradient 2 has B as origin. (see Plot 2)



Plot 2


4. y = ¹/₂·x + 6 meets the x-axis means that y = 0
then, y = ¹/₂·x + 6 = 0 ⇒ x = − 12
if another line passes through C, its equation should be written y = ²/₃·x + 6 (see Plot 3)
The second equation colud the be written: ²/₃·x − y + 6 = 0, with a = ²/₃, b = −, c = 6.



Plot 3



5. Just do it like the previous exercises !!!




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6. Just do it like the previous exercises !!!

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