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Advanced Ext 1 Topics

Learn how to arrange individuals in various ways using permutations and combinations. Explore different scenarios to understand the concepts better and solve probability problems related to permutations. Find the coefficient of terms in algebraic expansions and calculate the likelihood of specific outcomes.

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Advanced Ext 1 Topics

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  1. Advanced Ext 1 Topics Permutations and Combinations

  2. Ext 1 Question • In how many ways can six girls and two boys be arranged in a row if the boys are not allowed together? • Answer: • Number of arrangements without restriction = 8! • Number of arrangements with boys together = 7! x 2! (i.e. tie the two boys together in 2! Ways, leaving 7! arrangements)

  3. Ext 1 Question • Therefore, 8! – 7!2! = 30 240

  4. Example 1 • In how many ways can 6 girls and 3 boys be arranged in a row, if no boys are allowed to be next to each other? • This is not as obvious or straight forward as the Ext 1 example…

  5. Answer: • Firstly, place the 6 girls. • This can be done in 6! ways. GGGGGG

  6. Secondly, place a boy in one of the seven vacancies. • This can be done in 7 ways. GGGGGG

  7. Secondly, place a boy in one of the seven vacancies. • This can be done in 7 ways. GGBGGGG

  8. Thirdly, place the second boy. • This can be done in 6 ways. GGBGGGG

  9. Thirdly, place the second boy. • This can be done in 6 ways. GGBGGGBG

  10. Lastly, place the third boy. • This can be done in 5 ways. GGBGGGBG

  11. Lastly, place the third boy. • This can be done in 5 ways. BGGBGGGBG

  12. Lastly, place the third boy. • This can be done in 5 ways. BGGBGGGBG There is nothing left to do! 6!7.6.5 = 151 200 ways

  13. Example 2 • Find the number of ways in which a selection of four letters can be made from the letters of the word TOMORROW

  14. TOMORROW • There are 4 possibilities: • All 4 letters selected are different eg: TOMR • 2 are alike, the other 2 different eg: OORW • 2 are alike, the other 2 alike eg: OORR • 3 are alike, the other different eg: OOOR Investigate each, one at a time.

  15. TOMORROW • All 4 letters selected are different • 5C4 = 5 ways

  16. TOMORROW • 2 are alike, the other 2 different • If the first 2 are OO, then there are 4C2 ways to choose the rest ie 6 ways • If the first 2 are RR, then there are 4C2 ways to choose the rest ie 6 ways

  17. TOMORROW • 2 are alike, the other 2 alike • OO and RR can only be chosen 1 way!

  18. TOMORROW • 3 are alike, the other different • OOO is chosen, leaving one letter to be chosen from 4, which can happen in4 ways

  19. TOMORROW • Putting it all together • 5+6+6+1+4 = 22 • 22 different selections of 4 letters can be made from TOMORROW!

  20. Example 3 • Find the number of ways in which an arrangement of four letters can be made from the letters of the word TOMORROW NB: This question involves permutations of words with repeated letters

  21. TOMORROW • For each of the 4 classifications outlined in Example 2, these are the number of ways they can be permuted: • 1. Each of the 5 possibilities for “All letters different” can be permuted 4! ways. 5 x 4! = 120 ROMT TOMR ROMW

  22. TOMORROW • For each of the 4 classifications outlined in Example 2, these are the number of ways they can be permuted: • 2. Each of the 12 possibilities for “2 alike, 2 different” can be permuted 4!/2! ways. 12 x 4!/2! = 144 OOMR

  23. TOMORROW • For each of the 4 classifications outlined in Example 2, these are the number of ways they can be permuted: • 3. The one possibility for “2 same 2 same” can be permuted 4!/(2!2!) ways. 4!/(2!2!) = 6 OORR

  24. TOMORROW • For each of the 4 classifications outlined in Example 2, these are the number of ways they can be permuted: 4. Each of the 4 possibilities for “3 letters same 1 different” can be permuted 4!/3! ways. 4 x 4!/3! = 16 OOOW

  25. TOMORROW Therefore, 120 + 144 + 6 + 16 = 286

  26. Complete the Extension 2 Probability Sheet (Answers are attached to it). • Don’t go past this slide till you have completed Binomial Probability in HSC Course.

  27. Example 4 - Extension 2 binomial probability type problem (a) Find the coefficient of the term in p2q3r5 in the expansion of (p + q + r)10. (b) In an election there are three candidates: Paolo, Quincey and Rebekah. It has been established that 20% of voters favour Paulo, 10% favour Quincey and 70% favour Rebekah. Ten voters are asked which candidate they prefer. What is the probability that 2 would vote for Paulo, 3 for Quincey and 5 for Rebekah? (Answer to 4 decimal places) • Method 1 In the expansion (p + q + r)10 there are 10 factors of (p + q + r), each of which contributes either to a ‘p’ or ‘q’ or ‘r’ to each term of the product. The number of ways of getting 2 × p, 3 × q and 5 × r in the product can be found by choosing the 2 factors from the original 10 that each contribute to a ‘p’, then choosing 3 factors from the remaining 8 that each contribute a ‘q’; the remaining 5 factors then each contribute an ‘r’. Therefore, Number of Ways = 10C2 × 8C3 × 5C5 = 2520 ways. Method 2 How many ways can 2 × ‘p’, 3 × ‘q’ and 5 × ‘r’ be arranged? i.e. ppqqqrrrrr Number of arrangements = 10! / (2! 3! 5!) = 2520 ways Therefore, Coefficient of p2q3r5 is 2520.

  28. Example 4 - Extension 2 binomial probability type problem (b) In an election there are three candidates: Paolo, Quincey and Rebekah. It has been established that 20% of voters favour Paulo, 10% favour Quincey and 70% favour Rebekah. Ten voters are asked which candidate they prefer. What is the probability that 2 would vote for Paulo, 3 for Quincey and 5 for Rebekah? (Answer to 4 decimal places) (b) Let p = probability of voter favours Paulo = 0.2 Let q = probability of voter favours Quincey= 0.1 Let r = probability of voter favours Rebekah= 0.7 From (a): Required Probability = 2520p2q3r5 = 2520 × (0.2)2 × (0.1)3 × (0.7)5 = 0.0042 (to 4 decimal places)

  29. Also consider the Examples in the text (p238) – Example 13, 15

  30. HSC Questions • 2010 to current… see sheet.

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