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Combinations and The Binomial Theorem.

Combinations and The Binomial Theorem.

Hi! I am Aalam and i am going to be talking about Combinations and Binomial Theorem.

Combination: 

• An unordered collection of elements.
• In permutation we select and order elements. But with a combination we only select the element.
• Must use the formula. Dash Method cannot be used.

      Formula:
    

 

•  N = The total number of items in the sample.
•  R = The number of items to be selected from in the sample.
•  N > R 

Example: A class consists of 12 girls and 10 boys. A committee is to be selected consisting of 7         members. In how many ways can this be done if:

A) There are to be 4 boys and 4 girls on the committee?

4 Boys:      n = 10    Total number of boys.

                  r = 4       Number of boys that could be selected.

                  10C4 = 210 

3 Girls:      n = 12   Total number of girls.

                  r = 3      Number of girls that could be selected.

                  12C4  = 220 

Total:       210x220 = 46200

B) There are to be at least 5 men on the committee?

5 Boys:     n = 10    Total number of boys.

                 r = 5       At least 5 boys need to be selected.

                 10C5 = 252

2 Girls:    n = 12   Total number of girls.

                r = 2      As there are 5 boys and the committee consists of 7 people 2 girls could be                                           selected.

                  12C2 = 66

Total:      252x66 = 16632

OR

6 Boys:    n = 10    Total number of boys.

                r = 6       6 boys could be selected as the question says at least 5 boys.

               10C6 = 210

1 Girls:   n = 12     Total number of girls.

               r = 2        As there are 6 boys and the committee consists of 7 people 1 girls can be                                                selected.

               12C1 = 12

Total:     210x12 = 2520

OR

7 Boys     n = 10   Total number of boys.

                r = 7      7 boys could be selected as the question says at least 5 boys.

                10C7 = 120

0 Girls as the committee consists of 7 people and we already got 7 boys.

Total =   16632+2520+120 = 19272

C) It does not matter which sex is chosen?

n = 22   As there are 10 boys and 22 girls and sex doesnt matter there are a total of 22 people.

r = 7     Only 7 can be selected.

            22C7 = 170544

The Binomial Theorem:

      (a+b)ⁿ =  (_nC₀)(aⁿb⁰) + (_nC₁)(a^n-1)(b¹) + (_nC₂)(a^n-2)(b²) + … (_nC_n)( a⁰)(bⁿ)

• A binomial expansion where the exponent is n will have n+1 terms when expanded.
• A binomial expansion where the exponent is even will have an odd number of terms and will have a middle term when expanded. Ex. (x+y)^2
• A binomial expansion where the exponent is odd will have an even number of terms and will not have a middle term when expanded. Ex. (x+y)^3

    

Example: Expand and Simplify using the binomial expansion: (3x-y)⁵

a =  3x

b = -1y

n =  5

(3x-y)⁵   = ₅C₀ (3x)⁵(-y)⁰ + ₅C₁ (3x)⁴(-y)¹ + ₅C₂ (3x)³(-y)² + ₅C₃ (3x)²(-y)³ + ₅C₄ (3x)¹(-y)⁴ + ₅C₅                    (3x)⁰(-y)⁵

              

              = 1.243x⁵.1 + 5.81x⁴(-y) + 10. 27x³y² + 10.9x²(-y)³ + 5.3xy⁴ + 1. 1(-y)⁵

             

              =  243x⁵ – 405x⁴y + 270x³y³ – 90x²y³ + 15xy⁴ – y⁵

I hope you guys enjoyed this blog post have a great day!

I Choose!!

Permutations with Repetitions and Restrictions

Permutations: 

with Repetitions and Restrictions, and Case Restrictions

Hey class! I'm Ann and I'm going to talk about the things we covered today. HOW FUN!!!!!!

Anyways....

We covered two topics today, Permutations with Repetitions and Restrictions, and Permutations with Case Restrictions. 

Lets go back to our precious lessons. As we all know, permutation  is a set of distinct objects in an arrangement of objects, without repetition into a specific order. To permute a set of objects means to rearrange them.

We can use two methods to solve permutations problems:

• the permutation formula 
• Dash Method

Repetition means you are allowed to pick the same item more than once

Restrictions are when you are asked to place an item in a specific location

check out Carmelo and Billy`s blog post for more infos: http://precalculus40ssectioncwinter2015.blogspot.ca/2015/02/permutations-part-2-time-to-have-some.html 

Here are some EXAMPLES:

Permutations with Repetitions and Restrictions:

1)      Cathlene, JJ, PK, Aldrin, and Christy are going to be arranged for the Santa picture:

       a)If there is no restrictions on where they stand, how many arrangements are there

   n=5    r=5         

   5!= 120

       or

Dash method:

5• 4• 3• 2•1= 120

      b) If Christy (C) and PK (P) must be standing together, how many possibilities are there

2•1•3•2•1= 12 OR

CP

            3•2•1•2•1= 12 OR

                CP

            3•2•2•1•1= 12 OR           = 48

                    CP

            3•2•1• 2 •1= 12 OR

                       C  P

           

OR YOU CAN SOLVE IT THIS WAY:

CP= 1 entity = 2!       3 friends+CP= 4!

2!4!= 48

    c) If Pk and Christy cannot stand together, how many possibilities are there

          total arrangement- PK and Christy together= Christy and Pk not together

            120-48= 72

     d) What if JJ (J), Christy (C) and PK (P) must stand together, how many arrangements are possible

 JCP= 1 entity= 3!

2 friends= 2 entities

3!3!= 36

OR YOU CAN DO IT THIS WAY:

3•2•1•2•1=12 OR

JCP

2•3•2•1•1= 12 OR     =36

    JCP

2•1•3•2•1=12 

        JCP

Permutations with Case Restrictions:

1) How many different 4-digit numbers greater than 400 can you make using the digits 1, 2, 3, 4, 5 and 6. no digits are repeated

     1 •5 •4 •3= 60

     4

     1 •5 •4 •3= 60           =180

      5

    1 •5 •4 •3= 60

      6

OR YOU CAN DO IT THIS WAY:

3•5•4•3=180

4,5,6

I HOPE YOU ENJOYED MY BLOG POST. HAVE A GREAT DAY!

     

I CHOOSE!!!!!!! 

Permutations Part 2! 

Class Summary - February 5, 2015

In today's class, we covered two new topics: Factorial notation and Permutations (Part one).
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Factorial Notation - N! (Read as N factorial) means to multiply all of the positive integers from n consecutively down to 1.

          N = Whatever number is provided, must be an element of the positive integers
          n! = n(n-1)(n-2)(n-3)... 1

Examples: 

5! = 5 x 4 x 3 x 2 x 1                   = 120 

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

Exception:

0! = 1 because
n! = n(n-1)     - N cancels out on both sides, leaving (n-1). We now have to solve for N
n      n

n-1 = 0     - Move 1 to the other side
n  = +1


Any question that includes simplifying means to get rid of the factorial notation
Examples:

5! = 5 x 4 x 3 x 2 x 1      = 5
4! =       4 x 3 x 2 x 1

12! = 12 x 11 x 10 x 9 x 8!    = 12 x 11 x 10 x 9    =   11,880
8!   =  8!

Any question that includes solving means to solve for the variable, giving it a numerical value
Example:
Solve n(n-5)! = 24(n-6)!
- Expand question out
n(n-5)(n-6)!   = 24(n-6)          > n-6 cancels out on both sides
      (n-6)                  (n-6)

n(n-5)  = 24             > We're solving, so multiply n through the (n-5) bracket
n^2 - 5n = 24           > Move 24 to the other side to factor
n^2 - 5n -24 = 0       > Find two numbers that have a sum of -5 and product of -24 to factor
n^2 -8n +3n - 24 = 0
n(n-8)  3(n-8)
(n+3)(n-8)            

n = -3,  or n = +8    > Remember, we cannot take negative factorials. N cannot be negative when                                     placed into an equation. (N = -3) becomes an extraneous value; N is only equal                                     to +8.

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Permutation - Is a set of distinct objects in an arrangement of objects, without repetition into a specific order. To permute a set of objects means to rearrange them.
N = Amount of objects to permute
R = Amount of objects taken from N when you permute
      - Therefore, N must be bigger than R.    N > R

Formula: nPr =   - The formula can only be used if repetitions aren't allowed and there                                                       are no restrictions.


Example: How many three letter words composed from the 26 letters in the alphabet are possible if no repetitions are allowed?

N = 26      nPr =    26         = 26 x 25 x 24 x 23!      = 26 x 25 x 24    =   15600
R = 3                  (26-3)!                  23!


-  Solving a question gives N a numerical value.

Solve: nP1  = 20
N = N  
R = 1    

a)    n!    =  20
     (n-1)!    

 b) n(n-1)!    = 20    > (n-1)! cancels out
      (n-1)!

c)  n = 20

Welcome

Welcome to our blog. This space is designed for students of the Maples Collegiate, attending the Pre-Calculus 40S class, section C, with Mr.P. We are going to use this space to discuss our daily lessons, ask questions you didn't get a chance to ask in class, and to share your knowledge with other students. Most importantly we will use this blog to reflect on what we're learning.
Have a great semester.