Permutations with Some Identical Elements

1 / 7

# Permutations with Some Identical Elements - PowerPoint PPT Presentation

Permutations with Some Identical Elements. Sometimes you will deal with permutations in which some items are identical. Example How many different ways can you rewrite the word Like?. LIKE LIEK LKIE LKEI LEIK LEKI ILKE ILEK IELK IEKL IKLE IKEL

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Permutations with Some Identical Elements' - regis

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

### Permutations with Some Identical Elements

Sometimes you will deal with permutations in which some items are identical.

Example

How many different ways can you rewrite the word Like?

LIKE LIEK LKIE LKEI LEIK LEKI

ILKE ILEK IELK IEKL IKLE IKEL

KLIE KLEI KILE KIEL KELI KEIL

ELIK ELKI EILK EIKL EKLI EKIL

Therefore there are 24 different permutations 4P4= 4!

(Four spots with four different choices)

Try it with the word FALL

Keep track of the common letters by using L1 and L2

If the two L’s trade places, there are 2P2 = 2! ways that they can be arranged.

Therefore the total number of ways that the letters can be arranged is now expressed as

4! / 2! = 24 / 2 = 12

 Divide the total number of permutations by the number of ways that you can arrange the identical letters

Try it with the word SISS

From previous knowledge we can see that we take

4! / 3! Since 3! Is the number of ways that the S can be arranged.

= 24 / 6 = 4

The number of permutations of a set of n items of which a are identical is n! / a!

Example 2

Scott is a brick layer and wants to create a pattern on the side of a house. How many patterns could he make if he uses 5 red bricks, and one each of grey, white, green, and black?

 total number of bricks = 9  this becomes ‘n’ and a = 5 (identical bricks)

9! / 5! = 9 x 8 x 7 x 6 = 3024 patterns

Arrange the word ‘bookkeeper’

There are a total of 10 letters to choose from  10! choices if we didn’t care about the double/triple letters

10! / 2!2!3!  since there are 3 e’s, 2 0’s, 2 k’s

for a set of n objects containing a identical elements, b identical objects, c identical objects, there are

permutations.

Example,

Shawna is looking at all of her cars in the driveway. She has 3 mustangs, 2 motorcycles, 4 corvettes and 3 porsche’s. In how many ways can she arrange her cars?

There are 12 cars in total,

12! / (3!2!4!3!) = 277200

Homework

Pg 245 # 2,4ac, 7, 8a, 10, 12