Exercises on Combinations and Repetitions1

A - Nine cards, each of a different colour, are to be arranged in a line. How many different arrangements of the 9 cards are possible? The 9 cards include a pink card and a green card. How many different arrangements do not have the pink card next to the green card? Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line. How many different arrangements in total of 3 cards are possible? How many of the arrangements of 3 cards in part (iii) contain the pink card? How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card? Answers A - Answers and explanations Here we just need to multiply the series of descending natural numbers of 9, in other words using the factorial function. 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880 This issue can be solved by reversing the question like this: The total number of arrangements that do have the pink (P) card next to the green card (G), we assume both card are considered as one, attached together indeed! Therefore there are two possiblities: pink (P) card next to green card (G): PG = 8! = 40,320 green (G) card next to pink (P) card: GP = 8! = 40,320 >> P next to G = 2 × 8! = 80,640 Now working out the arrangements do not have the pink card next to the green card or P away from G P away from G = total number of arrangments − P next to G ==> P away from G = 9! − 2 × 8! = 362,880 − 80,640 = 282,240 Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line. How many different arrangements in total of 3 cards are possible? 9P3 = 9! (9 − 3)!  =  9! 6!  =  504 How many of the arrangements of 3 cards in part (iii) contain the pink card? With the restriction, there must be a pink (P) card, the two other cards are randomly selected and arranged from the 8 cards remaining. - 1 pink (P) out of 3 cards: ==> 3P1 = 3! 1!  =  3! = 6 ways or possibilities - 2 cards out of 8: ==> 8P2 = 8! (8 − 2)!  =  8! 6!  =  56 ways Then, there are 56 × 6 = 168 arrangements of 3 cards containing the pink card. How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card? Again, we can reverse the question: Firstly, we work out "The arrangements of 3 cards that do have the pink card next to the green card and green next to pink": PG or GP - 2 × 2! = 4 ways - 1 card randomly select out of the 7 remaining: 7 ways >> 4 × 7 = 28 ways P aways G = 504 (see (iii)) Therefore: The arrangements of 3 cards that do not have the pink card next to the green card is: 504 − 28 = 476

Exercises on Combinaitions and Repetitions

Combinations Dave is the Ceo of a committee. In how many ways can a committee of 5 be chosen from 10 people given, so that Dave ist one of them? a) 252 False! try again. b) 126 Yes! Well done. Explanation c) 495 False! try again. d) 3,024 False! try again. Mixed Exercises Determine whether each of the following situations is a Combination or Permutation. Determining how many different ways you can elect a Chairman and Co-Chairman of a committee if you have 10 people to choose from. Your choice: or Choosing the batting order on a baseball team with 13 players. How many different ways? Your choice: or Choosing 3 toppings for a pizza if there are 9 choices. Your choice: or Voting to allow 10 new members to join a club when there are 25 that would like to join. Your choice: or Creating an access code for a computer site using any 8 alphabet letters. Your choice: or Permutations A special type of password consists of four different letters of the alphabet, where each letter is used only once. How many different possible passwords are there? a) 14,950 False! try again. b) 358,800 Yes! Well done. Explanation c) 426 False! try again. d) 456,976 False! try again. How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive? a) 84 False! try again. b) 120 False! try again. c) 720 Yes! You got it. Explanation d) 504 False! try again. Dave is already chosen prior to the selection, so we need to choose another 4 from 9 remaining (no repetitions allow!). In choosing a committee, order doesn't matter; so we need the number of combinations of 4 people chosen from 9. 9C4 = 9! 4!(9 − 4)!  =  9! 4!·5!  =  9·8·7·6·5! 4!·5!  =  9·8·7·6 4·3·2·1  =  126 The number of permutations of 4 letters chosen from 26 is 26P4 = 26! (26 − 4)!  =  26! 22!  =  26·25·24·23·22! 22!  =  26·25·24·23 = 358,800 The number of permutations of 3 digits chosen from 10 is 10P3 = 10! (10 − 3)!  =  10! 7!  =  10·9·8·7! 7!  =  10·9·8 = 720 Top    |    More Exercises

Permutations

Permutations with repetition  Choosing from n things offers n choices each time! While choosing r of them, the permutations are: n · n · n ... (r times) Explanation: There are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multplying each time. We then use the exponent of r to write it down as follow: n · n · n ... (r times) = nr (n exponent of r) The Formula nr where: n is the number of things to choose from; and r the chosen things out of n (Here Repetition is allowed and order matters) Example: 3 numbers are chosen from 10 available (numbered 0, 1, 2, ..., 9)! What could be the permutations? The answer: 10 · 10 · ... (3 times) = 103 = 1.000 permutations Permutations without Repetition In this case, the number of available choices is reduced each time we pick a number. Example: Assuming we want to know what order 15 pool balls could be in! There're numbered 1, 2, ...., 15: see Img ➦ It goes like this: if we choose one number, let's say 5, we won't be able to choose aigain! Our next possibily of choosing will be amoung 14 poolballs: So, our first choice has 15 possibilities, the next choice 14 possibilities, then comes 13, 12, etc. And the total permutations will be: 15 · 14 · 13 · ... = 15! = 1.307.674.368.000 Now, we just want to choose 3 of them. So the permutation will be: 15 · 14 · 13 = 2.730 (Which means, there are 2.730 different ways to arrange 3 pool balls out of 15 balls) Without repetition the choices get reduced each time. To better express this mathematically, we use the Factorial function: symbolized !, which means multipying a series of descending natural numbers. Examples: 5! = 5 · 4 · 3 · 2 · 1 = 120 4! = 4 · 3 · 2 · 1 = 24 3! = 3 · 2 · 1 = 6 2! = 2 · 1 = 2 1! = 1 Note: 0! = 1 is an Axiom To recap: If we want to select all of the 15 pool balls, the permutations will be: 15! = 1.307.674.368.000 and If we want to select just 3, the permutations will be: 15! (15 − 3)!  =  15! 12!  =  15·14·13·12! 12!  =  15·14·13  =  2.730

Formula and Notations

P(n, r)

=

nPr

=

nPr

=

n! (n − r)!

With:

1. n is the number of things to choose from;

and


2. r the chosen things out of n
(No repetition, order matters)

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Combinations

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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

Solutions to Exercises on Vertex

1. Find (i) the vertex and (ii) the equation of the line of symetry of each of the following quadratic graphs.


1. y = 1(x -2)² + 3
   according to the formula, the quadratic equation is in the Vertex form:
   
   (i) its vertex is v(h, k), with h = -(-2) and k = 3 ==> v(2, 3)
   1 > 0, parabola opens upwards => Minimum
   
   (ii) its line of symetry is x = 2 (see grah)
   
   
   



The same method can be applied to b., c., d., e. and f.



7. y = 1(x -3)² + c
   
   (i) Vertex v(3, c)
   with c undefined constant, it acts like a parameter!!!
   1 > 0, parabola opens upwards => Minimum
   
   (ii) The line of symetry of the equation is x = c, c can take any value!


8. y = 1(x -p)² + q
   
   (i) h = p and k = q ==> vertex v(p, q)
   p, q constant values;
   1 > 0, parabola opens upwards => Minimum
   
   (ii) Line of symetry x = p, p constant.


9. y = (ax + b)² + c
   first bring the equation to a vertex form (put a as factor as in the vertex form): ==> y = a(x + ^b/_a) + c (vertex form)
   
   (i) h = −^b/_a, k = c, and the vertex: v(−^b/_a, c), a, b and c constant.
   a > 0, parabola opens upwards => Minimum
   
   (ii) Line of symetry x = −^b/_a, a and b constant.



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2. Find (i) the least (or, if appropriate, the greatest) value of each of the following quadratic expressions and (ii) the value of x for which this occurs.




Core Lesson



• Vertex form of a quadratic function: f(x) = a(x - h)² + k.
• a determines how the parabola opens:
  - a > 0, it opens upwards;
  - a < 0, it opens downwards.
• h and k are the values of the vertex.
• The greatest or least value of a quadratic function equals k.
• The greatest or least value of a quadratic function occurs at x = h.
• Vertex: v(h, k)


1. y = 1(x + 2)² - 1
   (x + 2)² - 1 is smallest for x + 2 = 0
   x + 2 = 0 when x = −2
   ==> the function will have its smallest output when x = −2
   
   so let's find the output now:
   f(-2) = (-2 + 2)² - 1 = (0)² −1 = −1
   
   So the least value of y = 1(x + 2)² - 1 is k = −1 and it occurs at x = −2, and the parabola opens upwards for a = 1 > 0.


2. y = 1(x - 1)² + 2
a = 1 > 0, the parabola opens upwards
==> its least value is k = 2 and it occurs at x = 1.


3. y = 5 - 1(x + 3)²
or y = −1(x + 3)² + 5
a = −1 < 0, the parabola opens downwards
==> its greatest value is k = 5 and it occurs at x = −3.


4. y = 1(2x + 1)² − 7
by factorizing 2: 2(x + ¹/₂)² − 7
a = 2 > 0, the parabola opens upwards
==> its least value is k = −7 and it occurs at x = ⁻¹/₂.


5. y = 3 − 2(x − 4)² or y = − 2(x − 4)² + 3
a = −2 < 0, the parabola opens downwards
==> the least value of the function is k = 3 and it occurs at x = 4.


6. y = 1(x + p)² + q, with p and q constant.
a = 1 > 0, the parabola opens upwards
==> the greatest value of the function is k = q and it occurs at x = −p.


7. y = 1(x − p)² − q, with p and q constant.
a = 1 > 0, the parabola opens upwards
==> the greatest value of the function is k = −q and it occurs at x = p.


8. y = r −1(x − t)² or y = −1(x − t)² + r, with t and r constant.
a = −1 < 0, the parabola opens downwards
==> the least value of the function is k = r and it occurs at x = t.


9. y = c −1(ax + b)² or y = −1(ax + b)² + c, with a, b and c constant.
factorize by a: −a(x + ^b/_a) + c (vertex form)
−a < 0, the parabola opens downwards
==> the least value of the function is k = c and it occurs at x = ^−b/_a.



3. Solve the following quadratic equations. Leave surds in your answer.


1. (x − 3)² − 3 = 0   | +3 (on both side)
(x − 3)² = 3   | √ (square root it)
x − 3 = ± √3   | +3
==> x = 3 ± √3


2. (x + 2)² − 4 = 0   | +4
(x + 2)² = 4   | √ (square root it)
x + 2 = ± 2
==> x = − 2 ± √2


3. 2(x + 3)² = 5   | ÷2
(x + 3)² = ⁵/₂   | √
x + 3 = ± √⁵/₂   | −3
==> x = − 3 ± √⁵/₂


4. (3x − 7)² = 8   | √
3x − 7 = ± √8   | +7
3x = 7 ± 2√2   | ÷3
==> x = ⁷/₃ ± ^2√2/₃


5. (x + p)² − q = 0   | +q
(x + p)² = q   | √
x + p = ± √q   | −p
==> x = − p ± √q


6. a(x + b)² − c = 0   | +c
a(x + b)² = c   | ÷a
(x + b)² = ^c/_a   | √
x + b = ± √^c/_a   | −b
==> x = − b ± √^c/_a.



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4. Express the following in completed square form


1. x² + 2x +2 (by completing the square)
(x² + 2x + 1) + 2 − 1
==> (x + 1)² + 1


2. x² −8x − 3 (completing the square)
(x² − 8x + 16)² − 3 − 16
==> (x − 4)² − 19


3. x² + 3x − 7 (completing the square)
(x² + 3x + 2.25)² − 7 − 2.25
==> (x + 1.5)² − 9.25


4. 5 − 6x + x², better x² − 6x + 5
(x² − 6x + 9) + 5 − 9
==> (x − 3)² − 4 or (x − 3)² − 2²


5. x² + 14x + 49 (This is a completed square Form)
==> (x + 7)²


6. 2x² + 12x − 5   |   ÷2 (divide by 2 to bring the equation to a regular quadratic form)
x² + 6x − ⁵/₂ (complete the square)
(x² + 6x + 9) − ⁵/₂ − 9
==> (x + 3)² − ²³/₂


7. 3x² − 12x + 3   |   ÷3
x² − 4x + 1 (complete the square)
(x² − 4x + 4) + 1 − 4
==> (x − 2)² − 3


8. 7 − 8x − 4x²   |   ÷4
<==> −x − 2x + ⁷/₄   |   ×(−1)
<==> x + 2x −⁷/₄   |   (complete the square)
(x² + 2x + 1) −⁷/₄ − 1
==> (x + 1) −¹¹/₄


9. 2x² + 5x − 3   |   ÷2
x² + ⁵/₂ − ³/₂   |   (complete the square)
(x² + ⁵/₂ + 1.56) − ³/₂ − 1.56
==> (x + 1.25) − 3.06




5. Use the completed square form to factorize the following expressions
1. x² − 2x − 35   |   (complete the square)
   
(x² − 2x + 1) − 35 − 1
(x − 1)² − 36 <==> (x − 1)² − 6²   |   √ (square it)
±(x + 1) ± 6
==> (x + 1 − 6)(x + 1 + 6)
==> (x − 5)(x + 7)


2. x² − 14x − 176   |   (complete the square)
   
(x² − 14 + 49) − 176 − 49 <==> (x − 7)² − 225 <==> (x − 7)² − 15²  |   √
==> (x − 7 + 15)(x − 7 − 15)
==> (x + 8)(x − 22)


3. x² + 6x − 432   |   (complete the square)
   
(x² + 6x + 9) − 432 − 9 <==> (x + 3)² − 441
(x + 3)² − 21²   |   √
==> (x + 3 − 21)(x + 3 + 21)
==> (x − 18)(x + 24)


4. 6x − 5x − 6   |   ÷6
   
x² − ^5x/₆ − 1   |   (complete the square)
(x² − ^5x/₆ + 0.18) − 1 − 0.18
(x − 0.42)² − 1.18 <==> (x − 0.42)² − ±1.08   |   √
==> (x − 0.42 − 1.09)(x − 0.42 + 1.09)
==> (x − 1.5)(x + 0.67)


5. 14 + 45x − 14x² <==> − 14x² + 45x + 14   |   − ÷14
x² − ^45x/₁₄ − 1   |   (complete the square)
(x² − 3.2x + 1.6²) − 1 − 1.6²
(x − 1.6)² − 3.5 <==> (x − 1.6)² − 1.9²   |   √
==> (x − 1.6 − 1.9)(x − 1.6 + 1.9)
==> (x − 3.5)(x − 0.3)


6. 12x² + x − 6   |   ÷12
x² − ^x/₁₂−¹/₂  |   (complete the square)
(x² − ^x/₁₂ + 0.041²) − 0.5 − 0.041²
(x − 0.041)² − 0.5 <==> (x − 0.041)² ± 0.7²
==> (x − 0.041 − 0.7)(x − 0.041 + 0.7)
==> (x − 0.74)(x + 0.66)



6. See Exercise 2

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Exercises on Vertex

BIS Berlin Pure Mathematics 1

Exercise 4A

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Vertex form of a quadratic equation

All we need:

1. the Vertex formula (V)
   Vertex: V(h, k)
   and
2. a quadratic equation in the Vertex form
   Quadratic equation: ax² + bx + c = 0
   The Vertex form: y = a(x - h)² + k

Properties:

• a > 0, parabola opens upwards => Minimum
• a < 0, parabola opens downwards => Maximum

Example:

1. y = 2(x - 1)² + 3
• Vertex: (1, 3)
• 2 > 0, parabola opens upwards (see graph)



2. y = - 3(x + 3)² + 4
• Vertex: (-3, 4)
• - 3 < 0, parabola opens downwards (see graph)

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Roots of quadratic equation with quadratic formula

Exercises:

1. 3x² + 5x + 1 = 0

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