(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.
(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.
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