Abstract image of a woman sitting on books and studying on a laptop

11X1 T06 02 permutations II (2010)

N
Nigel Simmons

The document discusses permutations of objects where some objects are the same. It provides examples of calculating the number of permutations for sets of objects with varying numbers of identical objects. The key points are: - The number of permutations of n objects with x identical objects is calculated as n! / x!. - This accounts for the number of ways to arrange the n objects overall divided by the number of ways to arrange the x identical objects. - Examples are given calculating the permutations of words that can be formed from letters where some letters are duplicated.

Education
1 of 35
Downloaded 32 times
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
Permutations Not All Different Case 3: Ordered Sets of n Objects,
Permutations Not All Different Case 3: Ordered Sets of n Objects, (i.e. some of the objects are the same)
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, (i.e. some of the objects are the same)
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different (i.e. some of the objects are the same) A B B A
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different (i.e. some of the objects are the same) A B B A 2! 2
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different A B C A C B B A C B C A C A B C B A
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different A B C A C B B A C B C A C A B C B A 3! 6
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different 2 same A B C A A B A C B A B A B A C B A A B C A C A B C B A 3! 6 3
2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different 2 same 3 same A B C A A B A A A A C B A B A B A C B A A B C A C A B C B A 3! 6 3 1
4 objects
4 objects all different A B C D C B A D A B D C C B D A A C B D C A B D A C D B C A D B A D B C C D B A A D C B C D A B B A C D D A C B B A D C D A B C B C A D D C A B B C D A D C B A B D A C D B A C B D A B D B A B
4 objects all different A B C D C B A D A B D C C B D A A C B D C A B D A C D B C A D B A D B C C D B A A D C B C D A B B A C D D A C B B A D C D A B C B C A D D C A B B C D A D C B A B D A C D B A C B D A B D B A B 4! 24
4 objects all different 2 same A B C D C B A D A A B C A B D C C B D A A A C B A C B D C A B D A B A C A C D B C A D B A B C A A D B C C D B A A C A B A D C B C D A B A C B A B A C D D A C B B A A C B A D C D A B C B A C A B C A D D C A B B C A A B C D A D C B A C A A B B D A C D B A C C A B A B D A B D B A B C B A A 4! 24 12
4 objects all different 2 same 3 same A B C D C B A D A A B C A A A B A B D C C B D A A A C B A A B A A C B D C A B D A B A C A B A A A C D B C A D B A B C A A D B C C D B A A C A B B A A A A D C B C D A B A C B A 4 B A C D D A C B B A A C B A D C D A B C B A C A B C A D D C A B B C A A B C D A D C B A C A A B B D A C D B A C C A B A B D A B D B A B C B A A 4! 24 12
4 objects all different 2 same 3 same A B C D C B A D A A B C A A A B A B D C C B D A A A C B A A B A A C B D C A B D A B A C A B A A A C D B C A D B A B C A A D B C C D B A A C A B B A A A A D C B C D A B A C B A 4 B A C D D A C B B A A C B A D C D A B C B A C A B C A D D C A B B C A A B C D A D C B A C A A B B D A C D B A C C A B A 4 same B D A B D B A B C B A A A A A A 4! 24 12 1
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are;
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; n! Number of Arrangements  x!
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x!
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ?
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3!
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3! 2! for the two O' s
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3! 2! for the two O' s 3! for the three N' s
If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3!  3326400 2! for the two O' s 3! for the three N' s
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible?
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions?
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2!
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of placing the vowels
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of Number of ways of placing the vowels placing the consonants
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of  1440 Number of ways of placing the vowels placing the consonants
2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of  1440 Number of ways of placing the vowels placing the consonants Exercise 10F; odd

More Related Content

PDF
12X1 T09 04 permutations II (2010)
Nigel Simmons
 
PDF
11X1 T06 03 combinations (2010)
Nigel Simmons
 
PDF
12X1 T09 05 combinations
Nigel Simmons
 
PDF
12X1 T09 05 combinations (2010)
Nigel Simmons
 
PPT
12X1 T09 04 permutations II
Nigel Simmons
 
PDF
12X1 T09 03 permutations I
Nigel Simmons
 
PDF
11 x1 t01 03 factorising (2014)
Nigel Simmons
 
PDF
11 x1 t05 02 permutations ii (2013)
Nigel Simmons
 
12X1 T09 04 permutations II (2010)
Nigel Simmons
 
11X1 T06 03 combinations (2010)
Nigel Simmons
 
12X1 T09 05 combinations
Nigel Simmons
 
12X1 T09 05 combinations (2010)
Nigel Simmons
 
12X1 T09 04 permutations II
Nigel Simmons
 
12X1 T09 03 permutations I
Nigel Simmons
 
11 x1 t01 03 factorising (2014)
Nigel Simmons
 
11 x1 t05 02 permutations ii (2013)
Nigel Simmons
 

Similar to 11X1 T06 02 permutations II (2010) (6)

PDF
11 x1 t05 02 permutations ii (2012)
Nigel Simmons
 
PDF
11X1 t05 03 combinations
Nigel Simmons
 
PDF
11 x1 t05 03 combinations (2013)
Nigel Simmons
 
PDF
11 x1 t05 03 combinations (2012)
Nigel Simmons
 
PDF
Formato respuestas 50 a e
Gabriel Estrada
 
PDF
Formato burbuja tareas
Gabriel Estrada
 
11 x1 t05 02 permutations ii (2012)
Nigel Simmons
 
11X1 t05 03 combinations
Nigel Simmons
 
11 x1 t05 03 combinations (2013)
Nigel Simmons
 
11 x1 t05 03 combinations (2012)
Nigel Simmons
 
Formato respuestas 50 a e
Gabriel Estrada
 
Formato burbuja tareas
Gabriel Estrada
 
Ad

More from Nigel Simmons (20)

PPT
Goodbye slideshare UPDATE
Nigel Simmons
 
PPT
Goodbye slideshare
Nigel Simmons
 
PDF
12 x1 t02 02 integrating exponentials (2014)
Nigel Simmons
 
PDF
11 x1 t01 02 binomial products (2014)
Nigel Simmons
 
PDF
12 x1 t02 01 differentiating exponentials (2014)
Nigel Simmons
 
PDF
11 x1 t01 01 algebra & indices (2014)
Nigel Simmons
 
PDF
12 x1 t01 03 integrating derivative on function (2013)
Nigel Simmons
 
PDF
12 x1 t01 02 differentiating logs (2013)
Nigel Simmons
 
PDF
12 x1 t01 01 log laws (2013)
Nigel Simmons
 
PDF
X2 t02 04 forming polynomials (2013)
Nigel Simmons
 
PDF
X2 t02 03 roots & coefficients (2013)
Nigel Simmons
 
PDF
X2 t02 02 multiple roots (2013)
Nigel Simmons
 
PDF
X2 t02 01 factorising complex expressions (2013)
Nigel Simmons
 
PDF
11 x1 t16 07 approximations (2013)
Nigel Simmons
 
PDF
11 x1 t16 06 derivative times function (2013)
Nigel Simmons
 
PDF
11 x1 t16 05 volumes (2013)
Nigel Simmons
 
PDF
11 x1 t16 04 areas (2013)
Nigel Simmons
 
PDF
11 x1 t16 03 indefinite integral (2013)
Nigel Simmons
 
PDF
11 x1 t16 02 definite integral (2013)
Nigel Simmons
 
PDF
11 x1 t16 01 area under curve (2013)
Nigel Simmons
 
Goodbye slideshare UPDATE
Nigel Simmons
 
Goodbye slideshare
Nigel Simmons
 
12 x1 t02 02 integrating exponentials (2014)
Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
Nigel Simmons
 
12 x1 t01 01 log laws (2013)
Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
Nigel Simmons
 
X2 t02 02 multiple roots (2013)
Nigel Simmons
 
X2 t02 01 factorising complex expressions (2013)
Nigel Simmons
 
11 x1 t16 07 approximations (2013)
Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
Nigel Simmons
 
11 x1 t16 05 volumes (2013)
Nigel Simmons
 
11 x1 t16 04 areas (2013)
Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
Nigel Simmons
 
11 x1 t16 01 area under curve (2013)
Nigel Simmons
 
Ad

Recently uploaded (20)

PPTX
Core Concepts of Personalized Learning and Virtual Learning Environments
Dr. Bushra Sumaiya
 
PDF
Mucosal Drug Delivery system_NDDS_BPHARMACY__SEM VII_PCI.pdf
SabsKhan1
 
PDF
MICROENCAPSULATION_NDDS_BPHARMACY__SEM VII_PCI .pdf
SabsKhan1
 
PDF
International_Financial_Reporting_Standa.pdf
VrindaTayade1
 
PDF
BP 505 T. PHARMACEUTICAL JURISPRUDENCE (UNIT 2).pdf
CMEmpire
 
PDF
advance database management system book.pdf
hinamannan364
 
PDF
CISA (Certified Information Systems Auditor) Domain-Wise Summary.pdf
infosecTrain
 
PPTX
A powerpoint presentation on the Revised K-10 Science Shaping Paper
JericEstaco2
 
PPTX
Share_Module_2_Power_conflict_and_negotiation.pptx
Lavanyaj33
 
PDF
Paper A Mock Exam 9_ Attempt review.pdf.
SameeraNaveed2
 
PDF
BP 505 T. PHARMACEUTICAL JURISPRUDENCE (UNIT 1).pdf
CMEmpire
 
PPTX
Virtual and Augmented Reality in Current Scenario
Shruti Dalela
 
DOCX
Cambridge-Practice-Tests-for-IELTS-12.docx
ssuser0d93ff
 
PDF
English Textual Question & Ans (12th Class).pdf
javaidpaswal33
 
PPTX
Computer Architecture Input Output Memory.pptx
Gunasundari Selvaraj
 
PPTX
Introduction to pro and eukaryotes and differences.pptx
AvaniKarumathil
 
PDF
FORM 1 BIOLOGY MIND MAPS and their schemes
arcade5
 
PDF
Race Reva University – Shaping Future Leaders in Artificial Intelligence
RACE REVA University
 
PDF
HVAC Specification 2024 according to central public works department
amitrobanipl
 
PDF
Uderstanding digital marketing and marketing stratergie for engaging the digi...
afreenpeshmam
 
Core Concepts of Personalized Learning and Virtual Learning Environments
Dr. Bushra Sumaiya
 
Mucosal Drug Delivery system_NDDS_BPHARMACY__SEM VII_PCI.pdf
SabsKhan1
 
MICROENCAPSULATION_NDDS_BPHARMACY__SEM VII_PCI .pdf
SabsKhan1
 
International_Financial_Reporting_Standa.pdf
VrindaTayade1
 
BP 505 T. PHARMACEUTICAL JURISPRUDENCE (UNIT 2).pdf
CMEmpire
 
advance database management system book.pdf
hinamannan364
 
CISA (Certified Information Systems Auditor) Domain-Wise Summary.pdf
infosecTrain
 
A powerpoint presentation on the Revised K-10 Science Shaping Paper
JericEstaco2
 
Share_Module_2_Power_conflict_and_negotiation.pptx
Lavanyaj33
 
Paper A Mock Exam 9_ Attempt review.pdf.
SameeraNaveed2
 
BP 505 T. PHARMACEUTICAL JURISPRUDENCE (UNIT 1).pdf
CMEmpire
 
Virtual and Augmented Reality in Current Scenario
Shruti Dalela
 
Cambridge-Practice-Tests-for-IELTS-12.docx
ssuser0d93ff
 
English Textual Question & Ans (12th Class).pdf
javaidpaswal33
 
Computer Architecture Input Output Memory.pptx
Gunasundari Selvaraj
 
Introduction to pro and eukaryotes and differences.pptx
AvaniKarumathil
 
FORM 1 BIOLOGY MIND MAPS and their schemes
arcade5
 
Race Reva University – Shaping Future Leaders in Artificial Intelligence
RACE REVA University
 
HVAC Specification 2024 according to central public works department
amitrobanipl
 
Uderstanding digital marketing and marketing stratergie for engaging the digi...
afreenpeshmam
 

11X1 T06 02 permutations II (2010)

  • 1. Permutations Not All Different Case 3: Ordered Sets of n Objects,
  • 2. Permutations Not All Different Case 3: Ordered Sets of n Objects, (i.e. some of the objects are the same)
  • 3. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, (i.e. some of the objects are the same)
  • 4. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different (i.e. some of the objects are the same) A B B A
  • 5. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different (i.e. some of the objects are the same) A B B A 2! 2
  • 6. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1
  • 7. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects
  • 8. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different A B C A C B B A C B C A C A B C B A
  • 9. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different A B C A C B B A C B C A C A B C B A 3! 6
  • 10. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different 2 same A B C A A B A C B A B A B A C B A A B C A C A B C B A 3! 6 3
  • 11. 2 objects Permutations Not All Different Case 3: Ordered Sets of n Objects, all different 2 same (i.e. some of the objects are the same) A B A A B A 2! 2 1 3 objects all different 2 same 3 same A B C A A B A A A A C B A B A B A C B A A B C A C A B C B A 3! 6 3 1
  • 13. 4 objects all different A B C D C B A D A B D C C B D A A C B D C A B D A C D B C A D B A D B C C D B A A D C B C D A B B A C D D A C B B A D C D A B C B C A D D C A B B C D A D C B A B D A C D B A C B D A B D B A B
  • 14. 4 objects all different A B C D C B A D A B D C C B D A A C B D C A B D A C D B C A D B A D B C C D B A A D C B C D A B B A C D D A C B B A D C D A B C B C A D D C A B B C D A D C B A B D A C D B A C B D A B D B A B 4! 24
  • 15. 4 objects all different 2 same A B C D C B A D A A B C A B D C C B D A A A C B A C B D C A B D A B A C A C D B C A D B A B C A A D B C C D B A A C A B A D C B C D A B A C B A B A C D D A C B B A A C B A D C D A B C B A C A B C A D D C A B B C A A B C D A D C B A C A A B B D A C D B A C C A B A B D A B D B A B C B A A 4! 24 12
  • 16. 4 objects all different 2 same 3 same A B C D C B A D A A B C A A A B A B D C C B D A A A C B A A B A A C B D C A B D A B A C A B A A A C D B C A D B A B C A A D B C C D B A A C A B B A A A A D C B C D A B A C B A 4 B A C D D A C B B A A C B A D C D A B C B A C A B C A D D C A B B C A A B C D A D C B A C A A B B D A C D B A C C A B A B D A B D B A B C B A A 4! 24 12
  • 17. 4 objects all different 2 same 3 same A B C D C B A D A A B C A A A B A B D C C B D A A A C B A A B A A C B D C A B D A B A C A B A A A C D B C A D B A B C A A D B C C D B A A C A B B A A A A D C B C D A B A C B A 4 B A C D D A C B B A A C B A D C D A B C B A C A B C A D D C A B B C A A B C D A D C B A C A A B B D A C D B A C C A B A 4 same B D A B D B A B C B A A A A A A 4! 24 12 1
  • 18. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are;
  • 19. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; n! Number of Arrangements  x!
  • 20. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x!
  • 21. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects
  • 22. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ?
  • 23. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3!
  • 24. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3! 2! for the two O' s
  • 25. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3! 2! for the two O' s 3! for the three N' s
  • 26. If we arrange n objects in a line, of which x are alike, the number of ways we could arrange them are; ways of arranging n! n objects Number of Arrangements  x! ways of arranging the like objects e.g. How many different words can be formed using all of the letters in the word CONNAUGHTON ? 11! Words  2!3!  3326400 2! for the two O' s 3! for the three N' s
  • 27. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible?
  • 28. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!
  • 29. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440
  • 30. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions?
  • 31. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2!
  • 32. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of placing the vowels
  • 33. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of Number of ways of placing the vowels placing the consonants
  • 34. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of  1440 Number of ways of placing the vowels placing the consonants
  • 35. 2001 Extension 1 HSC Q2c) The letters A, E, I, O and U are vowels (i) How many arrangements of the letters in the word ALGEBRAIC are possible? 9! Words  2!  181440 (ii) How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th, and 8th positions? 4! Words   5! 2! Number of ways of  1440 Number of ways of placing the vowels placing the consonants Exercise 10F; odd