1. **Problem statement:** How many ways can 6 people be lined up to get on a bus? 2. **Formula:** The number of ways to arrange $n$ distinct people in a line is $n!$. 3. **Calculation:** For 6 people, the number of ways is $$6! = 720.$$ 4. **Problem statement:** If a certain 2 persons refuse to follow each other, how many ways are possible? 5. **Explanation:** Total arrangements without restriction is $6! = 720$. 6. **Calculate arrangements where the 2 persons are together:** Treat the 2 persons as a single unit, so we have 5 units to arrange. 7. Number of ways to arrange these 5 units is $5! = 120$. 8. The 2 persons can be arranged among themselves in $2! = 2$ ways. 9. So, total arrangements with the 2 persons together is $5! \times 2! = 120 \times 2 = 240$. 10. **Number of arrangements where the 2 persons do NOT follow each other:** $$6! - 5! \times 2! = 720 - 240 = 480.$$ 11. **Problem statement:** How many arrangements can be made from the letters in A R T I C L E if: i. Consonants occupy the odd places. 12. **Letters:** A, R, T, I, C, L, E (7 letters) 13. **Consonants:** R, T, C, L (4 consonants) 14. **Vowels:** A, I, E (3 vowels) 15. **Positions:** 7 positions, odd places are positions 1, 3, 5, 7 (4 odd places) 16. Consonants must occupy these 4 odd places. 17. Number of ways to arrange 4 consonants in 4 odd places: $4! = 24$. 18. Number of ways to arrange 3 vowels in 3 even places: $3! = 6$. 19. Total arrangements: $$4! \times 3! = 24 \times 6 = 144.$$ 20. ii. R and T occupy the ends. 21. Ends are positions 1 and 7. 22. Number of ways to arrange R and T at ends: $2! = 2$. 23. Remaining 5 letters (A, I, C, L, E) can be arranged in the 5 middle positions: $5! = 120$. 24. Total arrangements: $$2! \times 5! = 2 \times 120 = 240.$$ 25. iii. T and C come together. 26. Treat T and C as a single unit. 27. Now we have 6 units: (TC), A, R, I, L, E. 28. Number of ways to arrange these 6 units: $6! = 720$. 29. T and C can be arranged among themselves in $2! = 2$ ways. 30. Total arrangements: $$6! \times 2! = 720 \times 2 = 1440.$$ 31. **Problem statement:** If you have four friends Ahmed Ali, Saleem, Qaiser, and Zeshaan, how many different ways can you text your friends if order matters? 32. Number of friends: 4 33. Number of ways to arrange 4 friends in order: $4! = 24$. 34. **Problem statement:** Out of 16 players, a team of 11 players is to be selected. How many teams can be formed if two particular players are to be included in each team? 35. Since 2 players must be included, select remaining 9 players from the other 14 players. 36. Number of ways: $$\binom{14}{9} = \frac{14!}{9!5!} = 2002.$$ 37. **Problem statement:** I don't quite remember my five-digit password. The first and last digits are the same, 542 is in the middle, and sum of password is divisible by 9. 38. Password format: $d_1 5 4 2 d_5$ with $d_1 = d_5$. 39. Sum of digits: $$d_1 + 5 + 4 + 2 + d_5 = 2d_1 + 11.$$ 40. This sum must be divisible by 9. 41. Let $S = 2d_1 + 11$. 42. Check values of $d_1$ from 0 to 9: - For $d_1=4$: $S=2\times4+11=19$ (not divisible by 9) - For $d_1=7$: $S=2\times7+11=25$ (not divisible by 9) - For $d_1=8$: $S=2\times8+11=27$ (divisible by 9) 43. So $d_1 = d_5 = 8$. 44. Password is: $$8 5 4 2 8.$$