(a) Five female and seven male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of employing the 4 teachers, if : (i) there is no restriction ; (ii) at least 2 of them are females.
(b) The table shows the positions awarded to 7 contestants by Judges X and Y in a competition.
| Contestant | P | Q | R | S | T | U | V |
| Judge X | 2 | 7 | 1 | 3 | 6 | 5 | 4 |
| Judge Y | 4 | 6 | 2 | 3 | 1 | 1 | 5 |
(i) Calculate, correct to one decimal place, the Spearman’s rank correlation coefficient.
(ii) Interpret your answer in b(i) above.
Explanation
(a) 5 female, 7 male = 12 in total.
(i) If no restriction, there are 12 teachers to fill up 4 vacancies.
Number of ways = (^{12}C_{4})
= (frac{12!}{4! 8!} = frac{12 times 11 times 10 times 9}{4 times 3 times 2})
= (495) ways.
(ii) No of ways if at least 2 of them are females
= No of ways [no restriction – 0 female – 1 female]
0 females implies 4 males
No of ways = (^{7}C_{4} = frac{7!}{4! 3!})
= (frac{7 times 6 times 5}{3 times 2} = 35) ways.
1 female = 1 female + 3 males
No of ways = (^{5}C_{1} times ^{7}C_{3})
= (frac{5!}{4! 1!} times frac{7!}{4! 3!})
= (5 times 35 = 175) ways.
(therefore) No of ways if at least 2 are females = 495 – (35 + 175) = 285 ways.
(b)
| X | Y | (R_{X}) | (R_{Y}) | (d = R_{X} – R_{Y}) | (d^{2}) |
| 2 | 4 | 6 | 3 | 3 | 9 |
| 7 | 6 | 1 | 1 | 0 | 0 |
| 1 | 2 | 7 | 5 | 2 | 4 |
| 3 | 3 | 5 | 4 | 1 | 1 |
| 6 | 1 | 2 | 6.5 | -4.5 | 20.25 |
| 5 | 1 | 3 | 6.5 | -3.5 | 12.25 |
| 4 | 5 | 4 | 2 | 2 | 4 |
| (sum) | 49.5 |
(R_{S} = 1 – frac{6 sum d^{2}}{n(n^{2} – 1)})
= (1 – frac{6(49.5)}{7(7^{2} – 1)})
= (1 – frac{297}{336})
= (frac{13}{112} = 0.116)
(ii) This implies that X and Y are fairly positively correlated. As X increases, Y increases as well.