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1 of 15
(Permutation and Combination)
• Suppose you have 6 different textbooks in
your backpack that you want to put on a
bookshelf. How many ways can the 6
books be arranged on the shelf?
Objectives
• Use organized lists and tree (branching)
diagrams to list all possible outcomes of a
trial.
• Identify whether permutation or
combination is appropriate to count the
number of outcomes of a trial.
• Use formulas or calculator commands to
evaluate permutation and combination
problems.
Counting Outcomes
• There are three principle ways to count all
the outcomes of a trial.
1. Draw a tree diagram
• Often a simple multiplication is enough
2. Systematically write all possibilities
3. Use permutation and combination techniques
Example: Tossing Coins
• Three coins are tossed (or one coin is
tossed three times) and the outcome of
heads or tails is observed.
• Draw a tree diagram (also called a
branching diagram) that shows all possible
outcomes.
• State a conclusion: How many equally
likely outcomes are there in this problem?
Tree Diagram for 3 Coins
• So there are 8 different equally likely outcomes.
H
T
First Toss Second Toss Third Toss
H
T
H
T
H
T
H
T
H
T
H
T
Write an Organized List
• For the coin-toss problem we just did,
write an organized list that shows all
possible outcomes (like HHH etc)
• Here is one possible organization
– HHH, HHT, HTH, HTT
– TTT, THT, TTH, THH
• There are other ways to organize
– Any method that is systematic (so that no
outcomes are missed) can work
Permutations of n objects
• Return to the bookshelf question.
• Suppose we change the problem to
arranging 10 books on the shelf. Now how
many arrangements are there?
• 10 x 9 x 8 x … = 3,628,800
– The shorthand notation for this is 10! (factorial)
• In general, there are n! ways to arrange n
objects
– This is called the permutation of n objects
– The key idea is that order matters
Arranging fewer than all the objects
• What if there are only 4 slots available on the bookshelf for the 10
books?
• Then there are 10 x 9 x 8 x 7 = 5040 ways to arrange 4 books out of
a group of 10
• Notice that this could be viewed as
• If we let n = the total number of objects and r = the number chosen
and arranged, then we could conclude that the number of ways to
arrange n objects taken r at a time is
• This can be quickly evaluated by (nPr)
– Try it now on your calculator with n = 10 and r = 4
10 9 8 7 6 5 4 3 2 1
6 5 4 3 2 1
        
    
!
( )!
n
n r

Example
• How many three-letter “words” can be
made from the letters A, B, C, and D?
– You can use your calculator to answer this.
– What are n and r in this problem?
• Don’t worry that many of them are not real words;
we don’t care in this context.
• Write an organized list of all the possible
“words”
– Be systematic; be sure you write them all.
Three-letter “words”
• Notice that being organized helps find all 24
permutations
• Notice also that ABC is different from ACB because in
permutation order matters
• Suppose we don’t care about order. Then we are
looking at combination, not permutation.
• How many combinations of three letters can be made
from an alphabet with four letters?
ABC ABD ACD BCD
ACB ADB ADC BDC
BCA BDA DCA CDB
BAC BAD DAC CBD
CAB DAB CAD DBC
CBA DBA CDA DCB
Three-letter “words”
• When order does not matter, ABC is the same as ACB (etc.), so
there are only 4 combinations in the 24 permutations.
– They can be seen in the 4 columns
• Notice that there are 3! (which is r!) permutations of each
combination.
– They can be seen in the 6 rows
• So to get the number of combinations of n objects taken r at a time,
divide permutations by r!
• The formula is
• The calculator command is (nCr) Try it now.
ABC ABD ACD BCD
ACB ADB ADC BDC
BCA BDA DCA CDB
BAC BAD DAC CBD
CAB DAB CAD DBC
CBA DBA CDA DCB
!
( )! !
n
n r r

Examples
• How many ways are there to form a 3
member subcommittee from a group of 12
people?
• How many ways are there to choose a
president, vice-president, and secretary
from a group of 12 people?
More Examples
• There are 5 cabins in the woods at a certain
vacation spot. Each cabin has a path that leads
to each of the other cabins. How many paths
are there in all?
– This is the combination of 5 things taken 2 at a time
(order does not matter), so 5C2=10
• There are 100 communications satellites orbiting
earth. Each satellite needs a transmit and
receive channel to talk to each of the other
satellites. How many channels are needed?
– This time AB is different from BA, so use permutation:
100P2=9900
• How many pentagons can be drawn from the
vertices of a regular 13-gon?
– Combination: 13C5=1287
Objectives
• Use organized lists and tree (branching)
diagrams to list all possible outcomes of a
trial.
• Identify whether permutation or
combination is appropriate to count the
number of outcomes of a trial.
• Use formulas or calculator commands to
evaluate permutation and combination
problems.

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Permutation and Combination excellant.ppt

  • 2. • Suppose you have 6 different textbooks in your backpack that you want to put on a bookshelf. How many ways can the 6 books be arranged on the shelf?
  • 3. Objectives • Use organized lists and tree (branching) diagrams to list all possible outcomes of a trial. • Identify whether permutation or combination is appropriate to count the number of outcomes of a trial. • Use formulas or calculator commands to evaluate permutation and combination problems.
  • 4. Counting Outcomes • There are three principle ways to count all the outcomes of a trial. 1. Draw a tree diagram • Often a simple multiplication is enough 2. Systematically write all possibilities 3. Use permutation and combination techniques
  • 5. Example: Tossing Coins • Three coins are tossed (or one coin is tossed three times) and the outcome of heads or tails is observed. • Draw a tree diagram (also called a branching diagram) that shows all possible outcomes. • State a conclusion: How many equally likely outcomes are there in this problem?
  • 6. Tree Diagram for 3 Coins • So there are 8 different equally likely outcomes. H T First Toss Second Toss Third Toss H T H T H T H T H T H T
  • 7. Write an Organized List • For the coin-toss problem we just did, write an organized list that shows all possible outcomes (like HHH etc) • Here is one possible organization – HHH, HHT, HTH, HTT – TTT, THT, TTH, THH • There are other ways to organize – Any method that is systematic (so that no outcomes are missed) can work
  • 8. Permutations of n objects • Return to the bookshelf question. • Suppose we change the problem to arranging 10 books on the shelf. Now how many arrangements are there? • 10 x 9 x 8 x … = 3,628,800 – The shorthand notation for this is 10! (factorial) • In general, there are n! ways to arrange n objects – This is called the permutation of n objects – The key idea is that order matters
  • 9. Arranging fewer than all the objects • What if there are only 4 slots available on the bookshelf for the 10 books? • Then there are 10 x 9 x 8 x 7 = 5040 ways to arrange 4 books out of a group of 10 • Notice that this could be viewed as • If we let n = the total number of objects and r = the number chosen and arranged, then we could conclude that the number of ways to arrange n objects taken r at a time is • This can be quickly evaluated by (nPr) – Try it now on your calculator with n = 10 and r = 4 10 9 8 7 6 5 4 3 2 1 6 5 4 3 2 1               ! ( )! n n r 
  • 10. Example • How many three-letter “words” can be made from the letters A, B, C, and D? – You can use your calculator to answer this. – What are n and r in this problem? • Don’t worry that many of them are not real words; we don’t care in this context. • Write an organized list of all the possible “words” – Be systematic; be sure you write them all.
  • 11. Three-letter “words” • Notice that being organized helps find all 24 permutations • Notice also that ABC is different from ACB because in permutation order matters • Suppose we don’t care about order. Then we are looking at combination, not permutation. • How many combinations of three letters can be made from an alphabet with four letters? ABC ABD ACD BCD ACB ADB ADC BDC BCA BDA DCA CDB BAC BAD DAC CBD CAB DAB CAD DBC CBA DBA CDA DCB
  • 12. Three-letter “words” • When order does not matter, ABC is the same as ACB (etc.), so there are only 4 combinations in the 24 permutations. – They can be seen in the 4 columns • Notice that there are 3! (which is r!) permutations of each combination. – They can be seen in the 6 rows • So to get the number of combinations of n objects taken r at a time, divide permutations by r! • The formula is • The calculator command is (nCr) Try it now. ABC ABD ACD BCD ACB ADB ADC BDC BCA BDA DCA CDB BAC BAD DAC CBD CAB DAB CAD DBC CBA DBA CDA DCB ! ( )! ! n n r r 
  • 13. Examples • How many ways are there to form a 3 member subcommittee from a group of 12 people? • How many ways are there to choose a president, vice-president, and secretary from a group of 12 people?
  • 14. More Examples • There are 5 cabins in the woods at a certain vacation spot. Each cabin has a path that leads to each of the other cabins. How many paths are there in all? – This is the combination of 5 things taken 2 at a time (order does not matter), so 5C2=10 • There are 100 communications satellites orbiting earth. Each satellite needs a transmit and receive channel to talk to each of the other satellites. How many channels are needed? – This time AB is different from BA, so use permutation: 100P2=9900 • How many pentagons can be drawn from the vertices of a regular 13-gon? – Combination: 13C5=1287
  • 15. Objectives • Use organized lists and tree (branching) diagrams to list all possible outcomes of a trial. • Identify whether permutation or combination is appropriate to count the number of outcomes of a trial. • Use formulas or calculator commands to evaluate permutation and combination problems.

Editor's Notes

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