(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.