Advanced Ext 1 Topics

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.