Simple Arrangements & Selections

1 / 16

# Simple Arrangements & Selections - PowerPoint PPT Presentation

Simple Arrangements & Selections Combinations & Permutations A permutation of n distinct objects is an arrangement , or ordering , of the n objects. An r -permutation of n distinct objects is an arrangement using r of the n objects.

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

## PowerPoint Slideshow about 'Simple Arrangements & Selections' - jana

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

### Simple Arrangements & Selections

Combinations & Permutations
• A permutation of n distinct objects is an arrangement, or ordering, of the n objects.
• An r-permutation of n distinct objects is an arrangement using r of the n objects.
• A r-combination of n objects is an unordered selection, or subset, of r of the n objects.
Notation
• P(n, r) denotes the number of r-permutations of n distinct objects.
• C(n, r) denotes the number of r-combinations of n distinct objects.
• It is spoken “n choose r”.
• The C(n, r) are known as “binomial coefficients” (for reasons that will become clear later).
Formula for P(n, r)
• By the product principle,
• P(n, 2) = n(n - 1), P(n, 3) = n(n - 1)(n - 2), …,
• P(n, n) = n!
• n choices for the 1st position
• (n - 1) choices for the 2nd position, …,
• 1 choice for the nth position.
• P(n, r) = n(n - 1) . . . (n - (r - 1))

= n(n - 1) . . . (n - (r - 1)) (n - r)!/(n - r)!

= n!/(n - r)!

Formula for C(n, r)
• All r-permutations can be counted by the product rule:

1. pick the r elements to be ordered: C(n, r)

2. Order the r elements: P(r, r) = r!

• That is, P(n, r) = C(n, r)P(r, r) = n!/(n - r)!
• Thus, C(n, r) = n!/(r!(n - r)!)
• C(n, r) = C(n, n - r).
• Give a counting argument for this.
Example

How many ways can 7 women & 3 men be arranged in a row, if the 3 men must be adjacent?

• Treat the men as a block.
• There are C(8, 1) ways to place the block of men among the women.
• There are P(3, 3) ways to arrange the men.
• There are P(7, 7) ways to arrange the women.
• By the product rule, there are (8)(3!)(7!) ways.
Example

How many ways are there to arrange the alphabet so that there are exactly 5 letters between a & b?

• Pick the position of the left letter of a & b. (This forces the position of the other letter.)
• Pick the order of a & b: 2.
• Arrange the other letters: P(24, 24).
• By the product rule: C(20, 1)(2)P(24, 24).
Example

How many 6-digit numbers without repetition are there so that the digits are nonzero, and 1 & 2 do not appear consecutively?

• It is simpler to do this indirectly:
• There are P(9, 6) ways to arrange 6 nonzero digits without repetition.
• Subtract the number of ways to arrange the digits so that 1 & 2 are consecutive:
• There are 2 ways to arrange 1 & 2.
• There are P(7,4) ways to arrange the other letters.
• By the addition rule, there are:

P(9, 6) - C(5, 1)(2)P(6, 4) ways to do this.

Example

How many ways are there to arrange the letters in the word MISSISSIPPI?

• Since the letters are not distinct, the answer is less than P(11,11) = 11!
• Pick the 1 position where the M goes: C(11,1)
• Pick the 4 positions where the Is go: C(10,4)
• Pick the 4 positions where the Ss go: C(6,4)
• Pick the 2 positions where the Ps go: C(2,2)
• There are C(11,1) C(10,4) C(6,4) C(2,2) ways.
Example
• A committee of 4 people is to be chosen from a set of 7 women and 4 men such that:
• There are at least 2 women?
• The committee has 2 of each sex, and Pres. & Mrs. Clinton cannot both be on the same committee?
There are at least 2 women?
• If we pick 2 women, then pick 2 more people without restriction, we get:
• C(7, 2)C(9,2).
• This is wrong; certain committees are counted more than once.
• See the tree diagram of this use of the product rule.
The Set Composition Rule
• When using the product rule, the elements of each component must be distinct.
• In the previous example, we cannot always tell which women were picked in the 1st phase, & which were picked in the 2nd phase.
To fix this: Use the addition rule
• The set of committees can be partitioned into:
• those with 2 women: C(7, 2)C(4, 2)
• those with 3 women: C(7, 3)C(4, 1)
• those with 4 women: C(7, 4)C(4, 0)

Thus, there are

C(7, 2)C(4, 2) + C(7, 3)C(4, 1) + C(7, 4)C(4, 0) such committees.

Example
• How many ways are there to arrange As & Us such that the 3rd U appears as the 12th letter in a 15 letter sequence?
• In the 1st 11 letters, U appears exactly 2 times.
• Using the product rule:
• Pick the positions of the 1st 2 Us: C(11, 2)
• Pick the 3 letters that follow the 12th: 23.
Characters
•    
•       
•   
•  
•     
•         