1. **Problem 1:** Find the set of possible values of $k$ such that the quadratic equation $$2x^2 + (3k - 2)x + k = 0$$ has real and distinct roots. 2. For a quadratic equation $ax^2 + bx + c = 0$, roots are real and distinct if and only if the discriminant $$\Delta = b^2 - 4ac > 0.$$ 3. Here, $a = 2$, $b = 3k - 2$, $c = k$. Calculate the discriminant: $$\Delta = (3k - 2)^2 - 4 \times 2 \times k = (3k - 2)^2 - 8k.$$ 4. Expand and simplify: $$ (3k - 2)^2 = 9k^2 - 12k + 4,$$ so $$\Delta = 9k^2 - 12k + 4 - 8k = 9k^2 - 20k + 4.$$ 5. Set the inequality for real and distinct roots: $$9k^2 - 20k + 4 > 0.$$ 6. Find the roots of the quadratic inequality boundary: Using quadratic formula, $$k = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 9 \times 4}}{2 \times 9} = \frac{20 \pm \sqrt{400 - 144}}{18} = \frac{20 \pm \sqrt{256}}{18}.$$ 7. Compute the roots: $$k = \frac{20 \pm 16}{18}.$$ So, $$k_1 = \frac{20 - 16}{18} = \frac{4}{18} = \frac{2}{9},$$ $$k_2 = \frac{20 + 16}{18} = \frac{36}{18} = 2.$$ 8. Since the quadratic $9k^2 - 20k + 4$ opens upwards (coefficient of $k^2$ is positive), the inequality $$9k^2 - 20k + 4 > 0$$ holds when $$k < \frac{2}{9} \text{ or } k > 2.$$ --- 9. **Problem 2 (a):** From 5 runners, 4 swimmers, and 3 gymnasts standing in a straight line, find the number of ways when all runners stand together, all swimmers stand together, and all gymnasts stand together. 10. Treat each group as a single block. There are 3 blocks which can be arranged in $$3! = 6$$ ways. 11. Inside each block, the individuals can be arranged among themselves: - Runners: $$5! = 120$$ ways - Swimmers: $$4! = 24$$ ways - Gymnasts: $$3! = 6$$ ways 12. Total number of ways is $$3! \times 5! \times 4! \times 3! = 6 \times 120 \times 24 \times 6 = 103680.$$ --- 13. **Problem 2 (b):** Select 4 members with at least one runner, one swimmer, and one gymnast. 14. Let the number selected from runners, swimmers, and gymnasts be $r, s, g$ respectively. We want $r + s + g = 4$, with $r \geq 1, s \geq 1, g \geq 1$. 15. Find all positive integer solutions: - (1,1,2) - (1,2,1) - (2,1,1) 16. Compute the number of ways for each case using combinations: - For (1,1,2): $$C(5,1) \times C(4,1) \times C(3,2) = 5 \times 4 \times 3 = 60$$ - For (1,2,1): $$5 \times C(4,2) \times 3 = 5 \times 6 \times 3 = 90$$ - For (2,1,1): $$C(5,2) \times 4 \times 3 = 10 \times 4 \times 3 = 120$$ 17. Total number of ways: $$60 + 90 + 120 = 270.$$