Question: Time Allowed: 1:30 hours. Answer all questions
QUESTION ONE
a. If $^nP_r = 506$ and $^nC_r = 253$, find the value of $r$ and $n$.
b. Prove that $^nP_r$ divided by $r! = ^nC_r$
c. Evaluate $^{n+1}C_{n-2}$ if $n = 15$
d. If $^{16}P_{r-1} : ^{15}P_{r-1} = 16 : 7$ then find $r$
e. If $^nP_r = 3024$ and $^nC_r = 126$, find the value of $r$ and $n$.
QUESTION TWO
The following data were collected on the market survey at Lumley
12, 8, 17, 13, 15, 9, 18, 11, 6, 1
Cal:
a) Mean deviation data from the mean
b) Variance
c) Standard deviation
d) Coefficient of standard deviation
graph shapes: no visible graph shapes or plotted graph present.
position_hint: centerUser: Slove with step by step formulas for me to understand
### Question One
1. **Problem Statement:**
Given $^nP_r = 506$ and $^nC_r = 253$, find $n$ and $r$.
2. **Formulas:**
- Permutation: $$^nP_r = \frac{n!}{(n-r)!}$$
- Combination: $$^nC_r = \frac{n!}{r!(n-r)!}$$
3. **Step a:**
From the problem, we have:
$$^nP_r = 506$$
$$^nC_r = 253$$
4. **Relate permutation and combination:**
$$^nP_r = r! \times ^nC_r$$
5. **Substitute values:**
$$506 = r! \times 253$$
6. **Solve for $r!$:**
$$r! = \frac{506}{253} = 2$$
7. **Find $r$ such that $r! = 2$:**
$$1! = 1, 2! = 2$$
So, $$r = 2$$
8. **Use combination formula to find $n$:**
$$^nC_2 = 253 = \frac{n(n-1)}{2}$$
9. **Multiply both sides by 2:**
$$n(n-1) = 506$$
10. **Solve quadratic:**
$$n^2 - n - 506 = 0$$
11. **Use quadratic formula:**
$$n = \frac{1 \pm \sqrt{1 + 4 \times 506}}{2} = \frac{1 \pm \sqrt{2025}}{2}$$
12. **Calculate square root:**
$$\sqrt{2025} = 45$$
13. **Find $n$ values:**
$$n = \frac{1 + 45}{2} = 23$$ or $$n = \frac{1 - 45}{2} = -22$$ (discard negative)
14. **Answer for part a:**
$$r = 2, n = 23$$
---
15. **Part b: Prove that $$\frac{^nP_r}{r!} = ^nC_r$$**
16. **Start with permutation:**
$$^nP_r = \frac{n!}{(n-r)!}$$
17. **Divide by $r!$:**
$$\frac{^nP_r}{r!} = \frac{n!}{r!(n-r)!}$$
18. **Recognize right side as combination:**
$$= ^nC_r$$
19. **Hence proved:**
$$\frac{^nP_r}{r!} = ^nC_r$$
---
20. **Part c: Evaluate $$^{n+1}C_{n-2}$$ for $n=15$**
21. **Substitute $n=15$:**
$$^{16}C_{13}$$
22. **Use combination formula:**
$$^{16}C_{13} = \frac{16!}{13!3!}$$
23. **Simplify factorial:**
$$= \frac{16 \times 15 \times 14 \times 13!}{13! \times 3!} = \frac{16 \times 15 \times 14}{3!}$$
24. **Calculate $3!$:**
$$3! = 6$$
25. **Calculate numerator:**
$$16 \times 15 = 240$$
$$240 \times 14 = 3360$$
26. **Divide:**
$$\frac{3360}{6} = 560$$
27. **Answer for part c:**
$$^{16}C_{13} = 560$$
---
28. **Part d: Given ratio $$^{16}P_{r-1} : ^{15}P_{r-1} = 16 : 7$$, find $r$**
29. **Write permutation formulas:**
$$^{16}P_{r-1} = \frac{16!}{(16-(r-1))!} = \frac{16!}{(17-r)!}$$
$$^{15}P_{r-1} = \frac{15!}{(15-(r-1))!} = \frac{15!}{(16-r)!}$$
30. **Form ratio:**
$$\frac{^{16}P_{r-1}}{^{15}P_{r-1}} = \frac{16!/(17-r)!}{15!/(16-r)!} = \frac{16!}{15!} \times \frac{(16-r)!}{(17-r)!}$$
31. **Simplify factorial ratio:**
$$\frac{16!}{15!} = 16$$
32. **Simplify factorial ratio:**
$$(17-r)! = (17-r) \times (16-r)!$$
33. **So ratio becomes:**
$$16 \times \frac{(16-r)!}{(17-r) \times (16-r)!} = \frac{16}{17-r}$$
34. **Given ratio equals $\frac{16}{17-r} = \frac{16}{7}$**
35. **Solve for $r$:**
$$17 - r = 7$$
$$r = 10$$
36. **Answer for part d:**
$$r = 10$$
---
37. **Part e: Given $^nP_r = 3024$ and $^nC_r = 126$, find $n$ and $r$**
38. **Use relation:**
$$^nP_r = r! \times ^nC_r$$
39. **Substitute values:**
$$3024 = r! \times 126$$
40. **Solve for $r!$:**
$$r! = \frac{3024}{126} = 24$$
41. **Find $r$ such that $r! = 24$:**
$$4! = 24$$
So, $$r = 4$$
42. **Use combination formula:**
$$^nC_4 = 126 = \frac{n(n-1)(n-2)(n-3)}{4!} = \frac{n(n-1)(n-2)(n-3)}{24}$$
43. **Multiply both sides by 24:**
$$n(n-1)(n-2)(n-3) = 3024$$
44. **Try integer values for $n$:**
For $n=9$:
$$9 \times 8 \times 7 \times 6 = 3024$$
45. **Answer for part e:**
$$r = 4, n = 9$$
---
### Question Two
Given data: $12, 8, 17, 13, 15, 9, 18, 11, 6, 1$
1. **Calculate mean ($\bar{x}$):**
$$\bar{x} = \frac{12 + 8 + 17 + 13 + 15 + 9 + 18 + 11 + 6 + 1}{10} = \frac{110}{10} = 11$$
2. **a) Mean deviation from mean:**
$$\text{Mean Deviation} = \frac{\sum |x_i - \bar{x}|}{n}$$
3. **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 absolute deviations:**
$$1+3+6+2+4+2+7+0+5+10 = 40$$
5. **Mean deviation:**
$$\frac{40}{10} = 4$$
6. **b) Variance ($\sigma^2$):**
$$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$$
7. **Calculate squared deviations:**
$$(12-11)^2=1, (8-11)^2=9, (17-11)^2=36, (13-11)^2=4, (15-11)^2=16, (9-11)^2=4, (18-11)^2=49, (11-11)^2=0, (6-11)^2=25, (1-11)^2=100$$
8. **Sum squared deviations:**
$$1+9+36+4+16+4+49+0+25+100 = 244$$
9. **Variance:**
$$\frac{244}{10} = 24.4$$
10. **c) Standard deviation ($\sigma$):**
$$\sigma = \sqrt{24.4} \approx 4.94$$
11. **d) Coefficient of standard deviation (C.V.):**
$$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 = \frac{4.94}{11} \times 100 \approx 44.91\%$$
---
**Final answers:**
- Question 1a: $r=2$, $n=23$
- Question 1b: $\frac{^nP_r}{r!} = ^nC_r$ proved
- Question 1c: $^{16}C_{13} = 560$
- Question 1d: $r=10$
- Question 1e: $r=4$, $n=9$
- Question 2a: Mean deviation = 4
- Question 2b: Variance = 24.4
- Question 2c: Standard deviation $\approx 4.94$
- Question 2d: Coefficient of standard deviation $\approx 44.91\%$