Combinations

Hello, my name is Jazztyn Ramos. I will be summarizing our lesson on Combinations.
We began our lesson by learning a new formula for Combinations and learning how we are going to be able to tell the differences in a question whether we use Combination or Permutation. 

Combinations: A combination is an unordered collection of elements.

We must be able to learn the differences between a Permutation and a Combination. One way we are able to learn this is by looking at a question and if we are able to see the words:

• Arranged: This will present a Permutation question
• Select/Choose: This will present a Combination question

In the booklet we are given 2 questions, 1 is giving us a Permutation question while the other gives us a Combination question.

1. How many ways can 5 books be arranged on a shelf?
 (This is a Permutation question because of the word Arranged)

We solve this question the way we would normally using Permutation:
                               5! = 120

2. How many ways can you select two books to read on your holiday if there are 5 to choose from?
(This is now a Combination question because of the word Select)

 Now in this question we are introduced to our first Combination question, this will give us the opportunity to use our new formula which is:

Combinations Formula

1. r will represent the number of books we are choosing: 2
2. n will represent the total number of books: 5
3. Now we plug it in to the formula:

                    nCr = 5C2 = 5!/2!(5-2)!
                                                       = 5!/2!3!
                                     = 20/2
                                     = 10

With a Permutation, we select and order the elements. With a Combination we only select the elements.
* WE MUST USE THE FORMULA FOR A COMBINATION, NOT THE DASH METHOD

On Example 4, its asks us to evaluate the following:

a) 7C5 = 7!/5!(7-5)! = 42/2 = 21
b) 7C2 = 7!/2!(7-2)! = 42/2 = 21

This Example introduces us to the formula: nCx = nCy
(This shows us that if there is the same number that represents "n" as long as the two numbers of "x" and "y" equal "n" we are able to get the same answer out of both equations.)

This rule is shown in Example 5: Solver for x.

a) xC6 = xC9 (To solve we add the two "r" to solve for "x")

6+9=15 (X = 15)

On Example 6 we are presented with a new way of using Combinations
Example 6: A class consists of 12 women and 10 men. A committee (Committee = Combination) is to be selected consisting of 7 members. In how many ways can this be done if

a) There are to be 4 men and 3 women in the committee?

(To answer this question we must use the Combination formula to chose from 4 men out of 10 and 3 women out of 12 so we use two Combination equations to figure out how we chose and then multiply both of the answers to get how many ways can 4 men and 3 women make up the committee)

4 men =  "r"                               3 women = "r"
the total of men 10 = "n"            the total of women 12 = "n"

10C4 = 210                                            12C3 = 220

Now we multiply 210 and 220 to figure out our answer which is = 46200 ways that 4 men and 3 women can make up the committee.

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

(Now this might be difficult but in this question we are given 3 possibilities to get our answer because even if it says that 5 men must be on the committee it does not restrict us from having all boys make up the committee or have at least 1 girl in it.)

Our first option is 5 men and 2 women:

First we chose the 5 men out of 10 which gives us the equation
10C5 = 252
Now we must look for 2 girls out of 12
12C2 = 66
Now we multiply 252 and 66 = 16632

Our second option is 6 men and 1 woman:

We chose 6 men out of 10
10C6 = 210
We are left with 1 woman out of 12
12C1 = 12
Now multiply 210 and 12= 2520

Our third and final option is all 7 men:

Chose 7 out of 10 men
10C7 = 120
Now we dont need any women out of 12
12C0 = 1
Multiply = 120

Now that we have the answers to the 3 options we are given we add them all up to see each way we are able to have 5 men in the committee

16632 + 2520 + 120 = 19272     There is a possibility of 19272 ways that a committee can be made                                                    up including 5 men

c) It does not matter which sex is chosen?

(This question is just basic, it just asks for us to chose 7 people to make up the committee out of the total of men and women which is 22)

22C7  = 22!/7!(22-7)!
         = 22*21*20*19*18*17*16*15!/7!15!
         = 22*21*20*19*18*17*16/7!
         =170544

This covers some of the things we have recently learned in the topic of Combinations. I hope that this post is able to help you guys out on this new topic and good luck to everyone.

                                                                                                                                                -Jazztyn