1. **Problem (a):** Find the number of distinct permutations of the letters of the word MATHEMATICS. - The word MATHEMATICS has 11 letters. - Count of each letter: M=2, A=2, T=2, H=1, E=1, I=1, C=1, S=1. - Number of distinct permutations is given by $$\frac{11!}{2!2!2!}$$. 2. **Solution (a):** $$11! = 39916800$$ $$2! = 2$$ So, $$\text{Permutations} = \frac{39916800}{2 \times 2 \times 2} = \frac{39916800}{8} = 4989600$$ --- 3. **Problem (b):** From 7 gentlemen principals and 4 lady principals, form a committee of 5 including at least one lady. - Total ways to select 5 from 11 = $$\binom{11}{5}$$. - Ways with no lady = select all 5 from 7 gentlemen = $$\binom{7}{5}$$. - Ways with at least one lady = total ways - no lady ways. 4. **Solution (b):** $$\binom{11}{5} = \frac{11!}{5!6!} = 462$$ $$\binom{7}{5} = \frac{7!}{5!2!} = 21$$ So, $$\text{Ways} = 462 - 21 = 441$$ --- 5. **Problem (c):** Supply function $$p = 0.4 e^{2x}$$, sales = 2000 units (x in thousands, so $$x=2$$). Find producer surplus. - Producer surplus = $$\int_0^x p \, dx - p(x) \times x$$. 6. **Solution (c):** Calculate $$\int_0^2 0.4 e^{2x} dx$$: $$\int 0.4 e^{2x} dx = 0.4 \times \frac{e^{2x}}{2} = 0.2 e^{2x}$$ Evaluate from 0 to 2: $$0.2 (e^{4} - 1)$$ Calculate $$p(2) \times 2 = 0.4 e^{4} \times 2 = 0.8 e^{4}$$ Producer surplus: $$0.2 (e^{4} - 1) - 0.8 e^{4} = 0.2 e^{4} - 0.2 - 0.8 e^{4} = -0.6 e^{4} - 0.2$$ Numerically, $$e^{4} \approx 54.598$$ So, $$-0.6 \times 54.598 - 0.2 = -32.7588 - 0.2 = -32.9588$$ Since surplus cannot be negative, re-check: Producer surplus is usually $$\int_0^x p(x) dx - p(x) x$$, but here supply function is price, so producer surplus = $$p(x) x - \int_0^x p(x) dx$$. Correcting: $$0.8 e^{4} - 0.2 (e^{4} - 1) = 0.8 e^{4} - 0.2 e^{4} + 0.2 = 0.6 e^{4} + 0.2$$ Numerically: $$0.6 \times 54.598 + 0.2 = 32.7588 + 0.2 = 32.9588$$ Producer surplus = 32.9588 (thousand units price units). --- 7. **Problem (d):** Sum of all natural numbers between 250 and 1000 divisible by 3. - First multiple of 3 after 250 is 252. - Last multiple of 3 before 1000 is 999. - Number of terms $$n = \frac{999 - 252}{3} + 1 = 250$$. 8. **Solution (d):** Sum of AP: $$S_n = \frac{n}{2} (a + l) = \frac{250}{2} (252 + 999) = 125 \times 1251 = 156375$$ --- 9. **Problem (e):** Demand law $$p = 20 - 2x - x^2$$, find consumer surplus at market demand $$x=3$$. - Consumer surplus = $$\int_0^3 p(x) dx - p(3) \times 3$$. 10. **Solution (e):** Calculate integral: $$\int_0^3 (20 - 2x - x^2) dx = [20x - x^2 - \frac{x^3}{3}]_0^3 = (20 \times 3 - 9 - 9) - 0 = 60 - 9 - 9 = 42$$ Calculate $$p(3) = 20 - 6 - 9 = 5$$ Consumer surplus: $$42 - 5 \times 3 = 42 - 15 = 27$$ --- 11. **Problem (f):** Find absolute max and min of $$f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 1$$ on $$[0,2]$$. 12. **Solution (f):** Find $$f'(x) = 12x^3 - 6x^2 - 12x + 6$$. Set $$f'(x) = 0$$: $$12x^3 - 6x^2 - 12x + 6 = 0 \Rightarrow 2x^3 - x^2 - 2x + 1 = 0$$ Try rational roots: x=1 $$2(1)^3 - 1 - 2 + 1 = 2 - 1 - 2 + 1 = 0$$ So, factor: $$(x-1)(2x^2 + x -1) = 0$$ Solve quadratic: $$2x^2 + x -1 = 0$$ $$x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ Roots: $$x = \frac{2}{4} = 0.5, \quad x = \frac{-4}{4} = -1$$ (discard -1 as outside interval) Critical points: $$x=0.5, 1$$ Evaluate $$f(x)$$ at 0, 0.5, 1, 2: $$f(0) = 1$$ $$f(0.5) = 3(0.5)^4 - 2(0.5)^3 - 6(0.5)^2 + 6(0.5) + 1 = 3(0.0625) - 2(0.125) - 6(0.25) + 3 + 1 = 0.1875 - 0.25 - 1.5 + 3 + 1 = 2.4375$$ $$f(1) = 3 - 2 - 6 + 6 + 1 = 2$$ $$f(2) = 3(16) - 2(8) - 6(4) + 12 + 1 = 48 - 16 - 24 + 12 + 1 = 21$$ Absolute minimum = 1 at $$x=0$$ Absolute maximum = 21 at $$x=2$$ --- 13. **Problem (g):** Number of ways to select 11 players from 12; (i) including captain, (ii) excluding captain. 14. **Solution (g):** Total ways: $$\binom{12}{11} = 12$$ (i) Include captain: select 10 from remaining 11: $$\binom{11}{10} = 11$$ (ii) Exclude captain: select all 11 from remaining 11: $$\binom{11}{11} = 1$$ --- 15. **Problem (h):** Find five numbers in AP with sum 35 and sum of squares 285. - Let numbers be $$a - 2d, a - d, a, a + d, a + 2d$$. Sum: $$5a = 35 \Rightarrow a = 7$$ Sum of squares: $$(a-2d)^2 + (a-d)^2 + a^2 + (a+d)^2 + (a+2d)^2 = 285$$ Calculate: $$5a^2 + 2d^2 + 8d^2 = 285$$ $$5(7^2) + 10 d^2 = 285$$ $$245 + 10 d^2 = 285 \Rightarrow 10 d^2 = 40 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2$$ Numbers: For $$d=2$$: 3, 5, 7, 9, 11 For $$d=-2$$: 11, 9, 7, 5, 3 --- **Final answers:** (a) 4989600 (b) 441 (c) 32.9588 (d) 156375 (e) 27 (f) Absolute max = 21 at x=2, Absolute min = 1 at x=0 (g) Total = 12, include captain = 11, exclude captain = 1 (h) 3, 5, 7, 9, 11