- By
**efrat** - Follow User

- 117 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' PERMUTATIONS AND COMBINATIONS' - efrat

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Now we’re going to do 3 books on a shelf again, but this time we’re going to choose them from a group of 8 books.

### If we were looking for different arrangements for all 8 books, then we would do 8!

A FACTORIAL is a counting method that uses consecutive whole numbers as factors.

The factorial symbol is !

Examples 5!= 5x4x3x2x1

= 120

7! = 7x6x5x4x3x2x1

= 5040

First, we’ll do some permutation problems.

Permutations are “arrangements”.

Let’s do a permutationproblem.

How many different arrangements are there for 3 books on a shelf?

Books A,B, and C can be arranged in these ways:

ABC ACB BAC BCA CAB CBA

Six arrangements or 3! = 3x2x1 = 6

In a permutation, the order of the books is important.

Each different permutation is a different arrangement.

The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.

1. How many ways can 4 books be arranged on a shelf?

4! or 4x3x2x1 or 24 arrangements

Here are the 24 different arrangements:

ABCD ABDC ACBD ACDB ADBC ADCB

BACD BADC BCAD BCDA BDAC BDCA

CABD CADB CBAD CBDA CDAB CDBA

DABC DACB DBAC DBCA DCAB DCBA

We’re going to have a lot more possibilities this time, because there are many groups of 3 books to be chosen from the total 8, and there are several different arrangements for each group of 3.

But we only want the different arrangements for groups of 3 out of 8, so we’ll do a partial factorial,

8x7x6

=336

1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there?

10x9x8x7x6 or 30240

2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there?

7x6x5x4 or 840

Now, we’ll do some combination problems.

Combinations are “selections”.

There are some problems where the orderof the items is NOT important.

These are called combinations.

You are just making selections, not making different arrangements.

Example: A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed?

Let’s call the 5 people A,B,C,D,and E.

Suppose the selected committee consists of students E, C, and A. If you re-arrange the names to C, A, and E, it’s still the same group of people. This is why the order is not important.

Because we’re not going to use all the possible combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees.

Therefore instead of using only 5x4x3, to get the fewer committees, we must divide.

(Always divide by the factorial of the number of digits on top of the fraction.)

Answer:

10 committees

5x4x3

3x2x1

1. How many possible committees of 2 people can be selected from a group of 8?

8x7

2x1

or 28 possible committees

2. How many committees of 4 students could be formed from a group of 12 people?

12x11x10x9

4x3x2x1

or 495 possible committees

When items are in a circular format, to find the number of different arrangements, divide:

n! / n

Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?

6! 6 = 120

Order Order is isimportant not important

Don’t Dodivide divide

Download Presentation

Connecting to Server..