Permutations Combinations 5839B0

### Question One 1. **Problem Statement:** a. Given $^nP_r = 506$ and $^nC_r = 253$, find $n$ and $r$. 2. **Recall formulas:** - Permutation: $$^nP_r = \frac{n!}{(n-r)!}$$ - Combination: $$^nC_r = \frac{n!}{r!(n-r)!}$$ 3. **Use the relation between permutation and combination:** $$^nP_r = r! \times ^nC_r$$ 4. **Given values:** $$^nP_r = 506, \quad ^nC_r = 253$$ 5. **Check if $^nP_r = r! \times ^nC_r$ holds:** $$506 = r! \times 253 \implies r! = \frac{506}{253} = 2$$ 6. **Since $r! = 2$, then $r = 2$** 7. **Find $n$ using $^nC_r = 253$:** $$^nC_2 = \frac{n(n-1)}{2} = 253$$ Multiply both sides by 2: $$n(n-1) = 506$$ 8. **Solve quadratic:** $$n^2 - n - 506 = 0$$ 9. **Use quadratic formula:** $$n = \frac{1 \pm \sqrt{1 + 4 \times 506}}{2} = \frac{1 \pm \sqrt{2025}}{2} = \frac{1 \pm 45}{2}$$ 10. **Positive root:** $$n = \frac{1 + 45}{2} = 23$$ --- 11. **Part b: Prove that $^nP_r / r! = ^nC_r$** 12. **Start with definitions:** $$^nP_r = \frac{n!}{(n-r)!}$$ $$^nC_r = \frac{n!}{r!(n-r)!}$$ 13. **Divide $^nP_r$ by $r!$:** $$\frac{^nP_r}{r!} = \frac{\frac{n!}{(n-r)!}}{r!} = \frac{n!}{r!(n-r)!} = ^nC_r$$ --- 14. **Part c: Evaluate $^{n+1}C_{n-2}$ for $n=15$** 15. Substitute $n=15$: $$^{16}C_{13} = ^{16}C_3$$ 16. Use combination formula: $$^{16}C_3 = \frac{16!}{3! \times 13!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1}$$ 17. Calculate numerator and denominator: $$= \frac{3360}{6} = 560$$ --- 18. **Part d: Given ratio $^{16}P_{r-1} : ^{15}P_{r-1} = 16 : 7$, find $r$** 19. Write permutations: $$^{16}P_{r-1} = \frac{16!}{(16-(r-1))!} = \frac{16!}{(17-r)!}$$ $$^{15}P_{r-1} = \frac{15!}{(15-(r-1))!} = \frac{15!}{(16-r)!}$$ 20. Form ratio: $$\frac{^{16}P_{r-1}}{^{15}P_{r-1}} = \frac{16!/(17-r)!}{15!/(16-r)!} = \frac{16!}{15!} \times \frac{(16-r)!}{(17-r)!}$$ 21. Simplify factorials: $$\frac{16!}{15!} = 16$$ $$\frac{(16-r)!}{(17-r)!} = \frac{1}{17-r}$$ 22. So ratio is: $$16 \times \frac{1}{17-r} = \frac{16}{17-r}$$ 23. Given ratio equals $\frac{16}{7}$: $$\frac{16}{17-r} = \frac{16}{7}$$ 24. Cross multiply: $$7 = 17 - r$$ 25. Solve for $r$: $$r = 17 - 7 = 10$$ --- 26. **Part e: Given $^nP_r = 3024$ and $^nC_r = 126$, find $n$ and $r$** 27. Use relation: $$^nP_r = r! \times ^nC_r$$ 28. Substitute values: $$3024 = r! \times 126 \implies r! = \frac{3024}{126} = 24$$ 29. Since $r! = 24$, then $r = 4$ (because $4! = 24$) 30. Use combination formula: $$^nC_4 = 126 = \frac{n(n-1)(n-2)(n-3)}{4!} = \frac{n(n-1)(n-2)(n-3)}{24}$$ 31. Multiply both sides by 24: $$n(n-1)(n-2)(n-3) = 3024$$ 32. Try integer values for $n$: - For $n=9$: $$9 \times 8 \times 7 \times 6 = 3024$$ 33. So $n=9$ and $r=4$ --- ### Question Two 1. **Data:** $12, 8, 17, 13, 15, 9, 18, 11, 6, 1$ 2. **Calculate mean:** $$\text{Mean} = \frac{12 + 8 + 17 + 13 + 15 + 9 + 18 + 11 + 6 + 1}{10} = \frac{110}{10} = 11$$ 3. **a) Mean deviation from mean:** Calculate absolute deviations: $$|12-11|=1, |8-11|=3, |17-11|=6, |13-11|=2, |15-11|=4, |9-11|=2, |18-11|=7, |11-11|=0, |6-11|=5, |1-11|=10$$ 4. Sum of deviations: $$1+3+6+2+4+2+7+0+5+10 = 40$$ 5. Mean deviation: $$\frac{40}{10} = 4$$ 6. **b) Variance:** Calculate squared deviations: $$(12-11)^2=1, (8-11)^2=9, (17-11)^2=36, (13-11)^2=4, (15-11)^2=16, (9-11)^2=4, (18-11)^2=49, (11-11)^2=0, (6-11)^2=25, (1-11)^2=100$$ 7. Sum of squared deviations: $$1+9+36+4+16+4+49+0+25+100 = 244$$ 8. Variance: $$\frac{244}{10} = 24.4$$ 9. **c) Standard deviation:** $$\sqrt{24.4} \approx 4.94$$ 10. **d) Coefficient of standard deviation:** $$\frac{\text{Standard deviation}}{\text{Mean}} \times 100 = \frac{4.94}{11} \times 100 \approx 44.91\%$$